RD Sharma 2022 Solutions for Class 9 Maths Chapter 6 - Factorization Of Polynomials

Answer:

(i) 3x² − 4x + 15 is a polynomial of degree 2.i.e Quadratic polynomial.

(ii) y² + 2$\sqrt{3}$ is a polynomial of degree 2 in y variable. i.e. Quadratic polynomial.

(iii) $3\sqrt{x}+\sqrt{2}x$

It is not a polynomial because exponent of x is $\frac{1}{2}$ which is not a positive integer.

(iv) $x-\frac{4}{x}$
It is not a polynomial because $\frac{4}{x}$ is fractional part.

(v) x¹² + y³ + t⁵⁰

It is a polynomial in three variables x, y and t.

(vi) ^{$1-\sqrt{5x}$}
Not polynomial, because the exponent of the variable of  ^{$1-\sqrt{5x}$}  is $\frac{1}{2},$ which is not a positive integer.

(vii) ^{$\frac{1}{x+1}$}
Not polynomial, $\frac{1}{x+1}$ is a rational expression $\left(f\left(x\right)=\frac{q\left(x\right)}{r\left(x\right)} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{expression}, r\left(x\right)\ne 0, q\left(x\right) \mathrm{and} r\left(x\right) \mathrm{are} \mathrm{polynomial}\right)$.

(viii) ^{$\frac{\left(x-2\right)\left(x-4\right)}{x}$}
Not polynomial, because the exponent of $\frac{\left(x-2\right)\left(x-4\right)}{x}=\frac{x^2-6x+8}{x}=x-6+8x^{-1}$ is –1, which is not a whole number.

(ix) ^{$\frac{1}{5x^{-2}}+5x+7=\frac{1}{5}{x}^2+5x+7$} is a polynomial of degree 2.i.e Quadratic polynomial.

Answer:

(i)

Coefficient of

(ii)

Coefficient of

(iii)

Coefficient of

(iv)

Coefficient of

Answer:

(i) 7x³ + 4x² − 3x + 12

Degree of the polynomial = 3

Because the highest power of x is 3.

(ii) 12 − x + 2x³

Degree of the polynomial = 3. Because the highest power of x is 3.

(iii) $5y-\sqrt{2}$

Degree of the polynomial = 1. Because the highest power of y is 1.

(iv) 7

Degree of the polynomial = 0. Because there is no variable term in the expression

(v) 0

Degree of the polynomial is not defined. As there is no variable or constant term

(vi) y³(1 – y⁴)
y³(1 – y⁴) = y³ – y⁷, degree is 7, because the maximum exponent of y is 7.

Answer:

(i)

The degree of the polynomial is 2. It is quadratic in x.

So, it is quadratic polynomial.

(ii)

The degree of the polynomial is 1. It is a linear polynomial in x.

(iii)

The degree of the polynomial is 1.

It is linear a polynomial in x.

(iv) 3y

The degree of the polynomial is 1. It is linear in y.

(v)

The degree of the polynomial is 2. It is quadratic polynomial in t.

(vi)

The degree of the polynomial is 4. Therefore, it is bi-quadratic polynomial in t.

Answer:

(i)

Here, x and y are two variables.

So, it is polynomial in two variables.

(ii)

Here, x and t are two variables.

So, it is polynomial in two variables.

(iii)

Here, only t is variable.

So, it is polynomial in one variable.

(iv)

Here, x, y and z are three variables

So, it is polynomial in three variables.

Answer:

(i)

It is cubic in x, so, it is cubic polynomial in x variable.

(ii)

Here, exponent of x in is not a positive integer, so, it is not a polynomial.

(iii)

It is a quadratic polynomial.

(iv) q(x) = 2x² − 3x + $\frac{4}{x}$+ 2

Here, exponent of x in is not a positive integer. So it is not a polynomial.

(v)

Here, exponent of x in x^3/2 is not a positive integer. So, it is not a polynomial.

(vi)

Here, exponent of x in is not a positive integer, so, it not a polynomial.

Answer:

(i)

The given expression is a Constant polynomial as there is no variable term in it.

(ii)

The given expression is Cubic polynomial as the highest exponent of x is 3.

(iii)

The given expression is linear polynomial as the highest exponent of x is 1.

(iv)

The given expression is Quadratic polynomial as the highest exponent of x is 2.

(v)

The given polynomial is an linear polynomial as the highest exponent of x is 1.

(vi)

The given polynomial is Cubic polynomial as the highest exponent of x is 3.

Answer:

An example of binomial of degree 35 is $f\left(t\right) = 4t^{35}-\frac{1}{2}$. It is a binomial as it has two terms and degree is 35 because highest exponent of t is 35.

An example of monomial of degree 100 is . It is a monomial as it has only one term and degree is 100 because highest exponent of x is 100

Answer:

Given: $\frac{x^3+2x+1}{5}-\frac{7}{2}{x}^2-x^6$
$=\frac{x^3}{5}+\frac{2x}{5} +\frac{1}{5}-\frac{7}{2}{x}^2-x^6$

(i) Degree of polynomial is the highest degree of its terms.
Here, degree of $\frac{x^3}{5}+\frac{2x}{5} +\frac{1}{5}-\frac{7}{2}{x}^2-x^6$  is 6.
(ii) Coefficient of x³ is $\frac{1}{5}$.
(iii) Coefficient of x⁶ is –1.
(iv) The constant term in the polynomial is $\frac{1}{5}$.

Answer:

(i) Coefficient of x² in $\frac{\mathrm{\pi }}{6}x+x^2-1$ is 1.
(ii) Coefficient of  x² in 3x – 5 is 0.
(iii)  (x – 1) (3x – 4) = 3x² – 7x + 4
 Coefficient of x² in 3x² – 7x + 4 is 3.
(iv) (2x – 5) (2x² – 3x + 1) = 4x³ – 6x² + 2x – 10x² + 15x – 5
= 4x³ – 16x² + 17x – 5
Coefficient of x² in 4x³ – 16x² + 17x – 5 is –16.

Answer:

(i) A monomial has a single term and degree 1 means the highest power of the variable is 1.

Monomial of degree 1 : 5x

(ii) A binomial has two terms and degree 20 means the highest power of the variable is 20.

Binomial of degree 20 : 6y²⁰ + 5

(iii) A tri-nomial has three terms and degree 2 means the highest power of the variable is 2.

Tri-nomial of degree 2 : x² + x + 1

Answer:

Let be the given polynomial

(i) The value of f (2) can be found by putting x = 2

        = 10 

(ii) The value of f (–3) can be found by putting x = –3

(iii) The value of f (0) can be found by putting x = 0

Answer:

(i) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is the zeros of the given polynomial.

(ii) To check whether the given number is the zero of the polynomial or not we have to find and

Now,

And ,

Hence, and x = 1 are the zeros of the polynomial.

(iii) To check whether the given number is the zero of the polynomial or not we have to find

Now,

And

Here, both the value of , are not satisfied the polynomial. Therefore, they are not the zeros of the polynomial.

(iv) To check whether the given number is the zero of the polynomial or not we have to find

And

Hence, x = 1, 2, 3 are the zeros of the polynomial p(x).

(v)To check whether the given number is the zero of the polynomial or not we have to find

Here, , does not satisfy therefore, is not a zero of the polynomial.

(vi)To check whether the given number is the zero of the polynomial or not we have to find

Hence, x = 0 is the zeros of the polynomial.

(vii) To check whether the given number is the zero of the polynomial or not we have to find

Hence, is zeros of the polynomial.

(viii)To check whether the given number is the zero of the polynomial or not we have to find

Hence, does not satisyf the polynomial, so, is not zero of the polynomial.

Answer:

The given polynomial is

If x = 2 is the root of the polynomial.

Then

Answer:

The given polynomial is

If is a zeros of the polynomial p(x).

then

Therefore,

Hence the value of

Answer:

The given polynomial is

f(x) =2x³ − 3x² + ax + b

If is zeros of the polynomial f(x), then f(0) = 0

Similarly, if x = − 1 is the zeros of the polynomial of,

Then, 

Putting the value of b from equation (1)

Thus,

Answer:

The given polynomial is

Here, f(x) is a polynomial with integer coefficient and the coefficient of highest degree term is 1. So, the integer roots of f(x) are factors of 6. Which are by observing.

          = 0   

Also,

And similarly,

f(−3) = 0

Therefore, the integer roots of the polynomial f(x) are −1, −2, − 3

Answer:

The given polynomial is

f(x) is a cubic polynomial with integer coefficients. If $\frac{b}{c}$ is rational root in lowest terms, then the values of b are limited
to the factors of 6 which are $\pm 1,\pm 2,\pm 3,\pm 6$ and the values of c are limited to the factor of 2 as $\pm 1,\pm 2$. Hence, the possible
rational roots are $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$. 

Since,

So, 2 is a root of the polynomial

Now, the polynomial can be written as,

Also,

Therefore,

Hence, the rational roots of the polynomialare 2, – 3/2 and – 1.

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show by actual division

So the remainder by actual division is 22

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will show remainder by actual division

So the remainder by actual division is −7

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is 92

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate the remainder by actual division

So the remainder by actual division is

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Now we will calculate remainder by actual division

So the remainder is

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

We will calculate remainder by actual division

So the remainder is 8

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is −3

Answer:

Let us denote the given polynomials as

We have to find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Remainder by actual division

Remainder is 0

Answer:

Let us denote the given polynomials as

(i) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(ii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iii) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(iv) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

(v) We will find the remainder whenis divided by.

By the remainder theorem, whenis divided bythe remainder is

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have,

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

By the given condition, the two remainders are same. Then we have, R₁ = R₂

Answer:

Given:  p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 
when p(x) is divided by (x + 1), remainder is 19.
By Remainder Theorem, we have
p(–1) = 19
⇒ (–1)⁴ – 2(–1)³ + 3(–1)² – a(–1) + 3a – 7 = 19
⇒ 1 + 2 + 3 + a + 3a – 7 = 19
⇒ 4a – 1 = 19
⇒4a = 20
⇒a = 5
Thus, the value of a is 5.

Therefore,  p(x) = x⁴ – 2x³ + 3x² – 5x + 8
Now, when p(x) is divided by (x + 2), remainder will be p(–2)
∴ p(–2) = (–2)⁴ – 2(–2)³ + 3(–2)² –5(–2) + 8
= 16 + 16 + 12 + 10 + 8
= 62

Hence, the remainder of polynomial p(x) when divided by (x + 2) is 62.

Answer:

Let us denote the given polynomials as

Now, we will find the remaindersandwhenandrespectively are divided by.

By the remainder theorem, whenis divided bythe remainder is

By the remainder theorem, whenis divided bythe remainder is

(i) By the given condition,

R₁ = R₂

(ii) By the given condition,

R₁ + R₂ = 0

(iii) By the given condition,

2R₁ − R₂ = 0

Answer:

Given that:

By the factor theorem,

If g(x) is a factor of f(x)

i.e.

Then

As is zero therefore g(x), is the factor of polynomial f(x).

Answer:

It is given that and 

By the factor theorem, g(x) is a factor of polynomial f(x)

i.e. 

Therefore,

$$\begin{gathered} f(-5)=3{\left(-5\right)}^4+17{\left(-5\right)}^3+9{\left(-5\right)}^2-7\left(-5\right)-10 \\ =3\times 625+17\times \left(-125\right)+225+35-10 \\ =1875-2125+250 \\ =0 \end{gathered}$$

Hence, g(x) is the factor of polynomial f(x).

Answer:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x).

i.e.

f (−3) = 0

Hence, g(x) is the factor of polynomial f (x).

Answer:

It is given that and

By the factor theorem, g(x) is the factor of polynomial f(x), if f (7) = 0.

Therefore, in order to prove that (x − 7) is a factor of f(x).

It is sufficient to show that f(7) = 0

Now,

Hence, (x − 7) is a factor of polynomial f(x).

Answer:

It is given that and

By the factor theorem,

(3x − 2) is the factor of f(x), if

Therefore,

In order to prove that (3x − 2) is a factor of f(x).

It is sufficient to show that

Now,

Hence, (3x − 2) is the factor of polynomial f(x).

Answer:

It is given that and

By factor theorem, (3 − 2x) is the factor of f(x), if = 0

Therefore,

In order to prove that (3 − 2x) is a factor of f(x). It is sufficient to show that

Now,

Hence, (3 − 2x), is the factor of polynomial f(x).

Answer:

It is given that and

We have

$\Rightarrow \left(x-2\right)$ and (x − 1) are factor of g(x) by the factor theorem.

To prove that (x − 2) and (x − 1) are the factor of f(x).

It is sufficient to show that f(2) and f(1) both are equal to zero.

And

Hence, g(x) is the factor of the polynomial f(x).

Answer:

Let be the given polynomial.

By factor theorem,

and are the factor of f(x).

If and f(4) are all equal to zero.

Now,

also

And

Hence, and are the factor of polynomial f(x).

Answer:

Let be the given polynomial.

By the factor theorem,

and are the factor of f(x).

If and f(7) are all equal to zero.

Therefore,

Also

And

Hence, and are the factor of the polynomial f(x).

Answer:

Let be the given polynomial.

By factor theorem, is the factor of f(x), if f (5) = 0

Therefore,

Hence, a = − 8.

Answer:

Let be the given polynomial.

By the factor theorem,

(x − 4) is a factor of f(x).

Therefore f(4) = 0

Hence ,

$$\begin{gathered} \Rightarrow 320-112-4a-28=0 \\ \Rightarrow 180-4a=0 \\ \Rightarrow a=\frac{180}{4}=45 \end{gathered}$$

Hence,

Answer:

Let 4x⁴ + 2x³ − 3x² + 8x + 5a be the polynomial.

By the factor theorem,

is a factor of f(x) if f(−2) = 0.

Therefore,

Hence,

Answer:

Let be the given polynomial.

By the factor theorem,

(x − 3) is a factor of f(x) if f (3) = 0

Therefore,

$$\begin{gathered} \Rightarrow 27k^2-9k+9k-k=0 \\ \Rightarrow 27k^2-k=0 \\ \Rightarrow k\left(27k-1\right)=0 \\ \Rightarrow k=0 or k=\frac{1}{27} \end{gathered}$$

Hence, the value of k is 0 or .

Answer:

(i) Let be the given polynomial.

By factor theorem, if (x − 2) is a factor of f(x), then f (2) = 0

Therefore,

Thus the value of a is 7/6.

(ii) Let f(x) = x⁵ − 3x⁴ − ax³ + 3ax² + 2ax + 4 be the given polynomial.

By the factor theorem, (x − 2) is a factor of f(x), if f (2) = 0

Therefore,

Thus, the value of a is .

Answer:

Let p(x) = x³ – 2mx² + 16
Since, p(x) is divisible by (x + 2).
∴ p(–2) = 0 (By remainder theorem)
⇒ (–2)³ – 2m(–2)² + 16 = 0
⇒ –8 – 8m + 16 = 0
⇒  8m = 8
⇒ m = 1

Hence, the value of m is 1.

Answer:

Let p(x) =  8x⁴ + 4x³ – 16x² + 10x + m and g(x) =  2x – 1
g(x) = 0
⇒ x = $\frac{1}{2}$
For g(x) to be a factor of p(x),
$p\left(\frac{1}{2}\right)=0$
$$\begin{gathered} \Rightarrow 8{\left(\frac{1}{2}\right)}^4+4{\left(\frac{1}{2}\right)}^3-16{\left(\frac{1}{2}\right)}^2+10\left(\frac{1}{2}\right)+m=0 \\ \Rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 1+1+m=0 \\ \Rightarrow m=-2 \end{gathered}$$

Hence, the value of m is –2.

Answer:

Given: p(x) = ax³ + x² – 2x + 4a – 9
g(x) = x + 1
⇒ x + 1 = 0
⇒ x = –1
If g(x) is a factor of p(x), then p(–1) = 0.
⇒ a(–1)³ + (–1)² – 2(–1) + 4a – 9 = 0
⇒ –a + 1 + 2 + 4a – 9 = 0
⇒ 3a – 6 = 0
⇒ a = 2

Hence, the value of a is 2.

Answer:

Let and be the given polynomial.

We have,

are the factors of g(x).

By factor theorem, if and both are the factor of f(x)

Then f(2) and f(−2) are equal to zero.

Therefore,

and

Adding these two equations, we get

Putting the value of a in equation (i), we get

Hence, the value of a and b are 1, − 8 respectively.

Answer:

Let be the given polynomial.

By the factor theorem, and are the factor of the polynomial f(x) if and both are equal to zero.

Therefore,

$$\begin{gathered} f(-1)=(-1{)}^3+3(-1{)}^2-2\alpha \left(-1\right)+\beta =0 \\ \Rightarrow f(-1)=-1+3+2\alpha +\beta =0 \\ \Rightarrow 2\alpha +\beta =-2 ...\left(\mathrm{i}\right) \end{gathered}$$

and

Subtracting (i) from (ii)

We get,

                                 $\alpha =-1$

Putting the value of $\alpha$ in equation (i), we get

Hence, the value of α and β are −1, 0 respectively.

Answer:

(i) Let be the given polynomial.

By factor theorem, (x − a) is a factor of the polynomial if f(a) = 0

Therefore,

Thus, the value of a is − 1.

(ii) Let be the given polynomial.

By factor theorem, (x − a) is a factor of f(x), if f(a) = 0.

Therefore,

Thus, the value of a is − 1/3.

Answer:

(i) Let be the given polynomial.

By the factor theorem, (+ a) is the factor of f(x), if f(− a) = 0, i.e.,

Thus, the value of a is − 4/3.

(ii) Let be the polynomial. By factor theorem, (x + a) is a factor of the f(x), if f(− a) = 0, i.e.,

Thus, the value of a is 0.

Answer:

Let and be the given polynomials.

We have,

Here, are the factor of g(x).

If f(x) is divisible by and , then and are factor of f(x).

Therefore, f(1) and f(−1) both must be equal to zero.

Therefore,

and

Adding both the equations, we get,

Putting this value in (i)

Hence, the value of p and q are 3, −3 respectively.

Answer:

Let be the given polynomial.

By factor theorem, and are the factors of f(x) if f(−1) and f(1) both are equal to zero.

Therefore,

and

Adding equation (i) and (ii), we get

Putting this value in equation (i), we get,

Hence, the value of a and b are – 2 and 2 respectively.

Answer:

Let and be the given polynomials.

We have,

Here, and are the factors of g(x),

Now,

By factor theorem,

and

Subtracting (ii) by (i), we get,

$$\begin{gathered} \left(2a-b\right)-\left(a-b\right)=-9-\left(-11\right) \\ a=2 \end{gathered}$$
Putting this value in equation (ii), we get,

Hence, the value of a and b are 2 and 13 respectively.

Answer:

Let be the given polynomial.

By factor theorem, if and both are factors of the polynomial f (x). if f(−1) and f(1) both are equal to zero.

Therefore,

And

Adding (i) and (ii), we get

And putting this value in equation (ii), we get,

a = 2

Hence, the value of a and b are 2 and −1 respectively.

Answer:

By divisible algorithm, when is divided by the reminder is a linear polynomial

Let be subtracted from p(x) so that the result is divisible by q(x).

Let

We have,

Clearly, and are factors of q(x), therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e. f (−4) and f (3) are equal to zero.

Therefore,

and

Adding (i) and (ii), we get,

Putting this value in equation (i), we get,

Hence, will be divisible by if 4 x − 4 is subtracted from it

Answer:

By division algorithm, when is divided by the reminder is a linear polynomial. So, let r(x) = ax + b be added to p(x) so that the result is divisible by q(x)

Let

We have,

$$\begin{gathered} q\left(x\right) = 3x^2 + 7x - 6 \\ = 3x^2 + 9x - 2x - 6 \\ = 3x\left(x + 3\right) - 2\left(x + 3\right) \\ = \left(3x - 2\right) \left(x + 3\right) \end{gathered}$$

Clearly, $\left(3x - 2\right)$ and $\left(x + 3\right)$ are factors of q(x).

Therefore, f(x) will be divisible by q(x) if and are factors of f(x), i.e.,

and f(−3) are equal to zero.

Now,

$$\begin{gathered} f\left(\frac{2}{3}\right) = 0 \\ \Rightarrow 3{\left(\frac{2}{3}\right)}^3 + {\left(\frac{2}{3}\right)}^2 + \left(a - 22\right)\left(\frac{2}{3}\right) + 9 + b = 0 \\ \Rightarrow 3\times \frac{8}{27} + \frac{4}{9} + \frac{2a}{3} - \frac{44}{3} + 9 + b = 0 \\ \Rightarrow \frac{8}{9} + \frac{4}{9} - \frac{44}{3}+ 9 + \frac{2a}{3} + b = 0 \\ \Rightarrow \frac{8 + 4 - 132 + 81 }{9} + \frac{2a}{3} + b = 0 \\ \Rightarrow -\frac{39}{9} + \frac{2a}{3} + b = 0 \\ \Rightarrow \frac{2a}{3} + b = \frac{13}{3} \\ \Rightarrow 2a + 3b = 13 ........\left(\mathrm{i}\right) \end{gathered}$$

And

$$\begin{gathered} f\left(-3\right) = 0 \\ \Rightarrow 3{\left(-3\right)}^3 + {\left(-3\right)}^2 + \left(a - 22\right)\left(-3\right) + 9 + b = 0 \\ \Rightarrow -81 + 9 - 3a + 66 + 9 + b = 0 \\ \Rightarrow -3a + b = -3 \\ \Rightarrow b = -3 + 3a .........\left(\mathrm{ii}\right) \end{gathered}$$

Hence, p(x) is divisible by q(x), if is added to it.

Answer:

Let f(x) = x³ + 6x² + 11x + 6 be the given polynomial.

Now, put the we get

Therefore, is a factor of f(x).

Now,
$f\left(x\right)=x^3+5x^2+x^2+5x+6x+6$

Hence, are the factors of f(x).

Answer:

Let be the given polynomial.

Now, put the x = -1, we get

$$\begin{gathered} f(-1)={\left(-1\right)}^3+2{\left(-1\right)}^2-\left(-1\right)-2 \\ =0 \end{gathered}$$

Therefore, is a factor of polynomial f(x).

Now, x³ + 2x² − x − 2 can be written as,
$f\left(x\right)=x^3+3x^2-x^2-3x+2x-2$

Hence, are the factors of the polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,
$f\left(x\right)=x^3-7x^2+x^2+10x-7x+10$

Hence, are the factors of the polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

$$\begin{gathered} f\left(1\right)={\left(1\right)}^4-7{\left(1\right)}^3+9{\left(1\right)}^2+7\left(1\right)-10 \\ =1-7+9+7-10=0 \end{gathered}$$

Therefore, is a factor of polynomial f(x).

Now,
$f\left(x\right)=x^4-x^3-6x^3+6x^2+3x^2-3x+10x-10$

Where

Putting we get

Therefore, is a factor of g(x).

Now,
$g\left(x\right)=x^3-7x^2+x^2-7x+10x+10$

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f (y).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f(y).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(y).

Now,

Hence are the factors of polynomial f(y).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Now,

Hence are the factors of polynomial f(y).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,
$f\left(x\right)=x^4-3x^3+x^3-3x^2-4x^2+12x-4x+12$

Where

Putting we get

Therefore, is the factor of g(x).

Now,
$g\left(x\right)=x^3-2x^2-x^2-6x+2x+12$

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Answer:

Let $f\left(x\right)=x^4+10x^3+35x^2+50x+24$be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Where

Putting we get

Therefore, is the factor of g(x).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Answer:

Let be the given polynomial.

Now, putting we get

Therefore, is a factor of polynomial f(x).

Now,

Where

Putting we get

Therefore, is a factor of g(x).

Now,

From equation (i) and (ii), we get

Hence are the factors of polynomial f(x).

Answer:

(i) Let be the given polynomial.

is a factor of the polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

(ii) Let be the given polynomial.

Therefore is a factor of the polynomial f(x).

Now,

Hence are the factors of polynomial f(x).

Answer:

A real number a is a zero (or root) of the polynomial f(x), if f (a) = 0

Answer:

Since is a zero of polynomial f(x).

Therefore

The value of a is .

Answer:

When the polynomial f(x), divided by the remainder will be

Thus, the reminder = 5

Answer:

When the polynomial f(x) = x³ + 4x² + 4x − 3 is divided by x the remainder will be.

Thus, the reminder = −3

Answer:

As is a factor of polynomial

i.e.

$$\begin{gathered} {\left(-1\right)}^3+a=0 \\ \Rightarrow a=1 \end{gathered}$$

Thus, the value of a = 1

Answer:

When polynomial divided by

The remainder is 6.

i.e.

Thus, the value of

Answer:

Polynomial with degree 2 is known as a quadratic polynomial.

Hence, 7 – 9x + 2x² is called a quadratic polynomial.

Answer:

Polynomial with degree 3 is known as a cubic polynomial.

​Hence, 12x – 7x² + 4 – 2x³ is called a cubic polynomial.

Answer:

Polynomial with degree 4 is known as a biquadratic polynomial.

​Hence, 12x – 7x² + 4 – 2x³ is called a biquadratic polynomial.

Answer:

Let f(x) = ax³ + x² – 2x + 4a – 9

It is given that one factor of f(x) is (x + 1).

Therefore, $f\left(x\right)=0 \mathrm{when} x=-1$.

On putting x = –1 in f(x) = 0, we get

$$\begin{gathered} a{\left(-1\right)}^3+{\left(-1\right)}^2-2\left(-1\right)+4a-9=0 \\ \Rightarrow -a+1+2+4a-9=0 \\ \Rightarrow 3a-6=0 \\ \Rightarrow 3a=6 \\ \Rightarrow a=2 \end{gathered}$$

Hence, if x + 1 is a factor of the polynomial ax³ + x² – 2x + 4a – 9, then a = 2.

Answer:

Let f(x) = 81x³ – 45x² +3a – 6

It is given that one factor of f(x) is (3x – 1).

Therefore, $f\left(x\right)=0 \mathrm{when} x=\frac{1}{3}$.

On putting x = $\frac{1}{3}$ in f(x) = 0, we get

$$\begin{gathered} 81{\left(\frac{1}{3}\right)}^3-45{\left(\frac{1}{3}\right)}^2+3a-6=0 \\ \Rightarrow 81\left(\frac{1}{27}\right)-45\left(\frac{1}{9}\right)+3a-6=0 \\ \Rightarrow 3-5+3a-6=0 \\ \Rightarrow 3a-8=0 \\ \Rightarrow 3a=8 \\ \Rightarrow a=\frac{8}{3} \end{gathered}$$

Hence, if 3x – 1 is a factor of the polynomial 81x³ – 45x² +3a – 6, then the value of a is $\underline{\frac{8}{3}}.$

Answer:

Let f(x) = x³ + x² – 9x – 9

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x,

we use remainder theorem, put x = 0.

f(0) is the remainder.

Now,

f(0) = (0)³ + (0)² – 9(0) – 9
       = –9

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x is –9.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1,

put x +1 = 0.

f(–1) is the remainder.

Now,

f(–1) = (–1)³ + (–1)² – 9(–1) – 9
         = –1 + 1 + 9 – 9
         = 0

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 1 is 0.

To find the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2,

put x +2 = 0.

f(–2) is the remainder.

Now,

f(–2) = (–2)³ + (–2)² – 9(–2) – 9
         = –8 + 4 + 18 – 9
         = 5

Hence, the remainder obtained when x³ + x² – 9x – 9 is divided by x + 2 is 5.


Hence, the remainders obtained when x³ + x² – 9x – 9 is divided by x, x + 1 and x + 2 respectively are –9, 0 and 5.

 

Answer:

Let f(x) = 4x³ – 3x² + 2x – 1

To find the remainder obtained when 4x³ – 3x² + 2x – 1 is divided by 2x + 1,

we use remainder theorem, put 2x + 1 = 0.

f($-\frac{1}{2}$) is the remainder.

Now,

$$\begin{gathered} f\left(-\frac{1}{2}\right)=4{\left(-\frac{1}{2}\right)}^3-3{\left(-\frac{1}{2}\right)}^2+2\left(-\frac{1}{2}\right)-1 \\ =4\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)-1-1 \\ =-\frac{1}{2}-\frac{3}{4}-2 \\ =\frac{-2-3-8}{4} \\ =-\frac{13}{4} \end{gathered}$$

Hence, the remainder when f(x) = 4x³ – 3x² + 2x – 1 is divided by 2x + 1 is $\underline{-\frac{13}{4}}.$

Answer:

Given:
Degree of a polynomial f(x) = 7
Degree of polynomial f(x) g(x) = 56

Degree of polynomial f(x) g(x) = Degree of â€‹polynomial f(x) × Degree of â€‹polynomial g(x)
⇒ 56 = 7 × Degree of â€‹polynomial g(x)
⇒ Degree of â€‹polynomial g(x) = $\frac{56}{7}$
⇒ Degree of â€‹polynomial g(x) = 8

Hence, degree of g(x) is 8.

Answer:

Let f(x) = x¹⁵

To find the remainder obtained when x¹⁵ is divided by x + 1,

we use remainder theorem, put x + 1 = 0.

f(−1) is the remainder.

Now,

$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^{15} \\ =-1 \end{gathered}$$

Hence, the remainder when x¹⁵ is divided by x + 1 is −1.

Answer:

Let p(x) =  x² – 4x + 3

$$\begin{gathered} p\left(2\right)={\left(2\right)}^2-4\left(2\right)+3 \\ =4-8+3 \\ =-1 \\ p\left(-1\right)={\left(-1\right)}^2-4\left(-1\right)+3 \\ =1+4+3 \\ =8 \\ p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-4\left(\frac{1}{2}\right)+3 \\ =\frac{1}{4}-2+3 \\ =\frac{1}{4}+1 \\ =\frac{1+4}{4} \\ =\frac{5}{4} \\ \mathrm{Now}, \\ p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=-1-8+\frac{5}{4} \\ =-9+\frac{5}{4} \\ =\frac{-36+5}{4} \\ =-\frac{31}{4} \end{gathered}$$

Hence, if $p\left(x\right)=x^2-4x+3, \mathrm{then} p\left(2\right)-p\left(-1\right)+p\left(\frac{1}{2}\right)=\underline{-\frac{31}{4}}.$