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#### Page No 52:

${x}^{2}-2x-8=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x+2x-8=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-4\right)+2\left(x-4\right)=0$

Sum of zeroes =  4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes =  (4)(3)=121=constant termcoefficient of x2

#### Page No 52:

${x}^{2}+7x+12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x+3x+12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+4\right)+3\left(x+4\right)=0$

Sum of zeroes =
Product of zeroes =

#### Page No 52:

Let f(x) = 2x x – 6

Hence, all the zeroes of the polynomial f(x) are .

Now,

Hence, the relationship between the zeros and the coefficients is verified.

#### Page No 52:

$2\sqrt{3}{x}^{2}-5x+\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2\sqrt{3}{x}^{2}-2x-3x+\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒2x\left(\sqrt{3}x-1\right)-\sqrt{3}\left(\sqrt{3}x-1\right)=0$

Sum of zeroes =  4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes =  (4)(3)=121=constant termcoefficient of x2

#### Page No 52:

Sum of zeroes =  4+(3)=71=(coefficient of x)(coefficient of x2)
Product of zeroes =  (4)(3)=121=constant termcoefficient of x2

#### Page No 52:

Let f(x) = 3x x – 4

Hence, all the zeroes of the polynomial f(x) are .

Now,

Hence, the relationship between the zeros and the coefficients is verified.

#### Page No 52:

Let f(x) = 5x+ 10x

Hence, all the zeroes of the polynomial f(x) are .

Now,

Hence, the relationship between the zeros and the coefficients is verified.

#### Page No 52:

Let α and β be the zeroes of the polynomial $p\left(x\right)=2{x}^{2}+5x+k$.

Now, using (1)

Hence, the value of k is 2.

#### Page No 52:

Given: $\left(x+a\right)$ is a factor of $2{x}^{2}+2ax+5x+10$
So, we have
$x+a=0\phantom{\rule{0ex}{0ex}}⇒x=-a$
Now, It will satisfy  the above polynomial.
Therefore, we will get
$2{\left(-a\right)}^{2}+2a\left(-a\right)+5\left(-a\right)+10=0\phantom{\rule{0ex}{0ex}}⇒2{a}^{2}-2{a}^{2}-5a+10=0\phantom{\rule{0ex}{0ex}}⇒-5a=-10\phantom{\rule{0ex}{0ex}}⇒a=2$

#### Page No 53:

Given: $a{x}^{2}+7x+b=0$
Since, $x=\frac{2}{3}$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get

Since, $x=-3$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get

From (1) and (2), we get

#### Page No 60:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes =  and Product of zeroes =

#### Page No 61:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =

$\because {\left(\frac{1}{\alpha }-\frac{1}{\beta }\right)}^{2}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\alpha }-\frac{1}{\beta }=±\frac{3}{2}$

#### Page No 61:

By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes =
$\therefore a-b+a+a+b=\frac{-\left(-3\right)}{1}\phantom{\rule{0ex}{0ex}}⇒3a=3\phantom{\rule{0ex}{0ex}}⇒a=1$

Now, Product of zeros =

#### Page No 63:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
${x}^{3}-\left(a+b+c\right){x}^{2}+\left(ab+bc+ca\right)x-abc$                                  ...(1)
Let
Substituting the values in (1), we get
${x}^{3}-\left(2-3+4\right){x}^{2}+\left(-6-12+8\right)x-\left(-24\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}-3{x}^{2}-10x+24$

#### Page No 63:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
${x}^{3}-\left(a+b+c\right){x}^{2}+\left(ab+bc+ca\right)x-abc$                                ...(1)
Let
Substituting the values in (1), we get
${x}^{3}-\left(\frac{1}{2}+1-3\right){x}^{2}+\left(\frac{1}{2}-3-\frac{3}{2}\right)x-\left(\frac{-3}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{3}-\left(\frac{-3}{2}\right){x}^{2}-4x+\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒2{x}^{3}+3{x}^{2}-8x+3$

#### Page No 63:

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x3 −(Sum of the zeroes)x2 + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
${x}^{3}-5{x}^{2}-2x+24$

#### Page No 63:

Quotient  $q\left(x\right)=x-3$
Remainder  $r\left(x\right)=7x-9$

#### Page No 63:

Quotient  $q\left(x\right)={x}^{2}+x-3$
Remainder  $r\left(x\right)=8$

#### Page No 63:

We can write

and

Quotient  $q\left(x\right)=-{x}^{2}-2$
Remainder  $r\left(x\right)=-5x+10$

#### Page No 63:

Let $f\left(x\right)=2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$ and $g\left(x\right)={x}^{2}-3$

Quotient  $q\left(x\right)=2{x}^{2}+3x+4$
Remainder  $r\left(x\right)=0$
Since, the remainder is 0.
Hence, ${x}^{2}-3$ is a factor of $2{x}^{4}+3{x}^{3}-2{x}^{2}-9x-12$

#### Page No 63:

Let f(x) = x4 + 2x+ 8x+ 12x + 18

It is given that when f(x) is divisible by x2 + 5, the remainder comes out to be px + q.

On division, we get the quotient x2 + 2x + 3 and the remainder 2x + 3.

Since, the remainder comes out to be px + q.

Therefore, p = 2 and q = 3.

Hence, the values of p and q are 2 and 3 respectively.

#### Page No 63:

By using division rule, we have
Divided = Quotient × Divisor + Remainder
$\therefore 3{x}^{3}+{x}^{2}+2x+5=\left(3x-5\right)g\left(x\right)+9x+10\phantom{\rule{0ex}{0ex}}⇒3{x}^{3}+{x}^{2}+2x+5-9x-10=\left(3x-5\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒3{x}^{3}+{x}^{2}-7x-5=\left(3x-5\right)g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒g\left(x\right)=\frac{3{x}^{3}+{x}^{2}-7x-5}{3x-5}$

$\therefore g\left(x\right)={x}^{2}+2x+1$

#### Page No 63:

We can write and

Quotient = $2x+3$
Remainder = $x+2$

By using division rule, we have

Divided = Quotient × Divisor + Remainder

#### Page No 63:

Let f(x) = 2x4 – 5x3 – 11x2  + 20x + 12

It is given that 2 and –2 are two zeroes of f(x)

Thus,  f(x) is completely divisible by (x + 2) and (x – 2).

Therefore, one factor of f(x) is (x2 – 4).

We get another factor of f(x) by dividing it with (x2 – 4).

On division, we get the quotient 2x2 – 5x  – 3.

Hence, all the zeroes of the polynomial f(x) are

#### Page No 64:

Let f(x) = xx– 14x– 2x + 24

It is given that  are two zeroes of f(x)

Thus,  f(x) is completely divisible by (x + $\sqrt{2}$) and (x – $\sqrt{2}$).

Therefore, one factor of f(x) is (x2 – 2).

We get another factor of f(x) by dividing it with (x2 – 2).

On division, we get the quotient x2 + x  – 12.

Hence, all the zeroes of the polynomial f(x) are

#### Page No 64:

Let f(x) = 2x– 13x+ 19x+ 7x – 3

It is given that  are two zeroes of f(x)

Thus,  f(x) is completely divisible by (x $-2-\sqrt{3}$) and (x – $2+\sqrt{3}$).

Therefore, one factor of f(x) is $\left({\left(x-2\right)}^{2}-3\right)$
$⇒$one factor of f(x) is (x2 – 4x + 1)

We get another factor of f(x) by dividing it with (x2 – 4x + 1).

On division, we get the quotient 2x2 – 5x  – 3.

Hence, all the zeroes of the polynomial f(x) are

#### Page No 64:

Let f(x) = 3x3 + 16x2 + 15x – 18

It is given that one of its zeroes is $\frac{2}{3}$.

Therefore, one factor of f(x) is (x – $\frac{2}{3}$).

We get another factor of f(x) by dividing it with (x – $\frac{2}{3}$).

On division, we get the quotient 3x+ 18x + 27.

Hence, other zero of the polynomial f(x) is –3.

#### Page No 64:

Let f(x) = 2x– 3x– 3x+ 6x – 2

It is given that 1 and $\frac{1}{2}$ are two zeroes of f(x).

Thus,  f(x) is completely divisible by (x – 1) and (x – $\frac{1}{2}$).

Therefore, one factor of f(x) is $\left(x-1\right)\left(x-\frac{1}{2}\right)$
$⇒$one factor of f(x) is $\left({x}^{2}-\frac{3}{2}x+\frac{1}{2}\right)$

We get another factor of f(x) by dividing it with $\left({x}^{2}-\frac{3}{2}x+\frac{1}{2}\right)$.

On division, we get the quotient 2x2 – 4.

Hence, all the zeroes of the polynomial f(x) are

#### Page No 65:

Let the other zeroes of ${x}^{2}-4x+1$ be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes =
$\therefore 2+\sqrt{3}+a=\frac{-\left(-4\right)}{1}\phantom{\rule{0ex}{0ex}}⇒a=2-\sqrt{3}$
Hence, the other zeroes of ${x}^{2}-4x+1$ is $2-\sqrt{3}$.

#### Page No 65:

$f\left(x\right)={x}^{2}+x-p\left(p+1\right)$
By adding and subtracting px, we get
$f\left(x\right)={x}^{2}+px+x-px-p\left(p+1\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(p+1\right)x-px-p\left(p+1\right)\phantom{\rule{0ex}{0ex}}=x\left[x+\left(p+1\right)\right]-p\left[x+\left(p+1\right)\right]$

So, the zeros of f(x) are −(p + 1) and p.

#### Page No 65:

$f\left(x\right)={x}^{2}-3x-m\left(m+3\right)$
By adding and subtracting mx, we get
$f\left(x\right)={x}^{2}-mx-3x+mx-m\left(m+3\right)\phantom{\rule{0ex}{0ex}}=x\left[x-\left(m+3\right)\right]+m\left[x-\left(m+3\right)\right]\phantom{\rule{0ex}{0ex}}=\left[x-\left(m+3\right)\right]\left(x+m\right)$

So, the zeros of f(x) are −m and m + 3.

#### Page No 66:

If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x2 − (α + β)x + αβ                .....(1)
Substituting the values in (1), we get
x2 − 6x + 4

#### Page No 66:

Given: x = 2 is one zero of the quadratic polynomial kx2 + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$k{\left(2\right)}^{2}+3\left(2\right)+k=0\phantom{\rule{0ex}{0ex}}⇒4k+6+k=0\phantom{\rule{0ex}{0ex}}⇒5k+6=0\phantom{\rule{0ex}{0ex}}⇒k=-\frac{6}{5}$

#### Page No 66:

Given: x = 3 is one zero of the polynomial 2x2 + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$2{\left(3\right)}^{2}+3+k=0\phantom{\rule{0ex}{0ex}}⇒21+k=0\phantom{\rule{0ex}{0ex}}⇒k=-21$

#### Page No 66:

Given: x = −4 is one zero of the polynomial x2x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
${\left(-4\right)}^{2}-\left(-4\right)-\left(2k+2\right)=0\phantom{\rule{0ex}{0ex}}⇒16+4-2k-2=0\phantom{\rule{0ex}{0ex}}⇒-2k=-18\phantom{\rule{0ex}{0ex}}⇒k=9$

#### Page No 66:

Given: x = 1 is one zero of the polynomial ax2 − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
$a{\left(1\right)}^{2}-3\left(a-1\right)1-1=0\phantom{\rule{0ex}{0ex}}⇒a-3a+3-1=0\phantom{\rule{0ex}{0ex}}⇒-2a=-2\phantom{\rule{0ex}{0ex}}⇒a=1$

#### Page No 66:

Given: x = −2 is one zero of the polynomial 3x2 + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
$3{\left(-2\right)}^{2}+4\left(-2\right)+2k=0\phantom{\rule{0ex}{0ex}}⇒12-8+2k=0\phantom{\rule{0ex}{0ex}}⇒k=-2$

#### Page No 66:

$f\left(x\right)={x}^{2}-x-6\phantom{\rule{0ex}{0ex}}={x}^{2}-3x+2x-6\phantom{\rule{0ex}{0ex}}=x\left(x-3\right)+2\left(x-3\right)$

So, the zeros of f(x) are 3 and −2.

#### Page No 66:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes =
$⇒1=\frac{-\left(-3\right)}{k}\phantom{\rule{0ex}{0ex}}⇒k=3$

#### Page No 66:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes =
$⇒3=\frac{k}{1}\phantom{\rule{0ex}{0ex}}⇒k=3$

#### Page No 66:

Given: (x + a) is a factor of 2x2 + 2ax + 5x + 10
We have
$x+a=0\phantom{\rule{0ex}{0ex}}⇒x=-a$
Since, (x + a) is a factor of 2x2 + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
$\therefore 2{\left(-a\right)}^{2}+2a\left(-a\right)+5\left(-a\right)+10=0\phantom{\rule{0ex}{0ex}}⇒-5a+10=0\phantom{\rule{0ex}{0ex}}⇒a=2$

#### Page No 66:

By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes =
$⇒a-b+a+a+b=\frac{-\left(-6\right)}{2}\phantom{\rule{0ex}{0ex}}⇒3a=3\phantom{\rule{0ex}{0ex}}⇒a=1$

#### Page No 66:

Equating x2x to 0 to find the zeros, we will get

Since,  x3 + x2ax + b is divisible by x2x.
Hence, the zeros of x2x will satisfy x3 + x2ax + b
$\therefore {\left(0\right)}^{3}+{0}^{2}-a\left(0\right)+b=0\phantom{\rule{0ex}{0ex}}⇒b=0\phantom{\rule{0ex}{0ex}}$

and

#### Page No 66:

By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =

#### Page No 66:

“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that

f(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).

#### Page No 66:

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x2 − (Sum of the zeros)x + Product of zeros
$⇒{x}^{2}-\left(-\frac{1}{2}\right)x+\left(-3\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{2}x-3\phantom{\rule{0ex}{0ex}}$
Hence, the required polynomial is ${x}^{2}+\frac{1}{2}x-3$.

x22x+13=03x232x+1=0

#### Page No 66:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
$\therefore f\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}-3=0\phantom{\rule{0ex}{0ex}}⇒3\left(2{x}^{2}-1\right)=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-1=0$
$⇒2{x}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{\sqrt{2}}$
Hence, the zeros of the quadratic polynomial f(x) = 6x2 − 3 are $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$.

#### Page No 66:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
$\therefore f\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4\sqrt{3}{x}^{2}+8x-3x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4x\left(\sqrt{3}x+2\right)-\sqrt{3}\left(\sqrt{3}x+2\right)=0$

Hence, the zeros of the quadratic polynomial $f\left(x\right)=4\sqrt{3}{x}^{2}+5x-2\sqrt{3}$ are .

#### Page No 66:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =

Solving αβ = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in $\alpha \beta =\frac{k}{1}$, we will get
k = 6

#### Page No 66:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = and Product of zeroes =

#### Page No 69:

(d) is the correct option.
A polynomial in x of degree n is an expression of the form p(x) =ao +a1x+a2x2 +...+an xn, where an $\ne$0.

#### Page No 69:

It is because in the second term, the degree of x is −1 and an expression with a negative degree is not a polynomial.

#### Page No 70:

Multiply by 10, we get
$10{x}^{2}-x-3$

#### Page No 75:

(b) 3,-1
Here, ${\mathrm{p}\left(\mathrm{x}\right)=x}^{2}-2x-3\phantom{\rule{0ex}{0ex}}$

#### Page No 75:

(a) −1
Here,

Comparing the given polynomial with , we get:

#### Page No 75:

(c) $\frac{2}{3}$
Here, $\mathrm{p}\left(x\right)={x}^{2}-2x+3k$
Comparing the given polynomial with $a{x}^{2}+bx+c$, we get:

It is given that
are the roots of the polynomial.

Also, =$\frac{c}{a}$

#### Page No 75:

(c) $\frac{5}{2}$
Let the zeroes of the polynomial be .
Here,
p
Comparing the given polynomial with $a{x}^{2}+bx+c$, we get:
a = 4, b = −8k and c = 9
Now, sum of the roots$=-\frac{b}{a}$

Here, p

#### Page No 75:

It is given that the two roots of the polynomial are 2 and −5.
Let
Now, sum of the zeroes, $\mathrm{\alpha }+\mathrm{\beta }$ = 2 + (5) = 3
Product of the zeroes, $\mathrm{\alpha \beta }$ = 2$×$5 = 10
∴ Required polynomial = ${x}^{2}-\left(\mathrm{\alpha }+\mathrm{\beta }\right)x+\mathrm{\alpha \beta }$
$={x}^{2}—\left(-3\right)x+\left(-10\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+3x-10$

#### Page No 75:

The given polynomial and its roots are .

#### Page No 75:

Let p$\left(\mathrm{x}\right)={x}^{3}+4{x}^{2}-3x-18$

#### Page No 75:

Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial =
$={x}^{2}-\left(-5\right)x+6\phantom{\rule{0ex}{0ex}}={x}^{2}+5x+6$

#### Page No 75:

Comparing the polynomial with ${x}^{3}-{x}^{2}\left(\alpha +\beta +\gamma \right)+x\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)-\alpha \beta \gamma$, we get: