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#### Page No 74:

(i) ${x}^{5}-2{x}^{3}+x+\sqrt{3}$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so, it is a polynomial of degree 5.

(ii) ${y}^{3}+\sqrt{3}y$ is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so, it is a polynomial of degree 3.

(iii) ${t}^{2}-\frac{2}{5}t+\sqrt{5}$ is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so, it is a polynomial of degree 2.

(iv) ${x}^{100}-1$ is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.

(v) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(vi) ${x}^{-2}+2{x}^{-1}+3$ is an expression having negative integral powers of x. So, it is not a polynomial.

(vii) Clearly, 1 is a constant polynomial of degree 0.

(viii) Clearly, $-\frac{3}{5}$ is a constant polynomial of degree 0.

(ix) $\frac{{x}^{2}}{2}-\frac{2}{{x}^{2}}=\frac{{x}^{2}}{2}-2{x}^{-2}$
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.

(x) $\sqrt[3]{2}{x}^{2}-8$ is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(xi) $\frac{1}{2{x}^{2}}=\frac{1}{2}{x}^{-2}$ is an expression having negative integral power of x. So, it is not a polynomial.

(xii) $\frac{1}{\sqrt{5}}{x}^{\frac{1}{2}}+1$
In this expression, the power of x is $\frac{1}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xiii) $\frac{3}{5}{x}^{2}-\frac{7}{3}x+9$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(xiv) ${x}^{4}-{x}^{\frac{3}{2}}+x-3$
In this expression, one of the powers of x is $\frac{3}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

(xv) $2{x}^{3}+3{x}^{2}+\sqrt{x}-1=2{x}^{3}+3{x}^{2}+{x}^{\frac{1}{2}}-1$
In this expression, one of the powers of x is $\frac{1}{2}$ which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.

#### Page No 75:

(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.

(ii) 6is a polynomial with degree 1. So, it is a linear polynomial.

(iii) –z3 is a polynomial with degree 3. So, it is a cubic polynomial.

(iv) 1 – y – y3 is a polynomial with degree 3. So, it is a cubic polynomial.

(v) x – x3 + x4 is a polynomial with degree 4. So, it is a quartic polynomial.

(vi) 1 + x + x2 is a polynomial with degree 2. So, it is a quadratic polynomial.

(vii) – 6x2 is a polynomial with degree 2. So, it is a quadratic polynomial.

(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.

(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.

#### Page No 75:

(i) The coefficient of x3 in $x+3{x}^{2}-5{x}^{3}+{x}^{4}$ is −5.

(ii) The coefficient of x in $\sqrt{3}-2\sqrt{2}x+6{x}^{2}$ is $-2\sqrt{2}$.

(iii) 2x – 3 + x= – 3 + 2x + 0xx3
The coefficient of x2 in 2x – 3 + x3 is 0.

(iv) The coefficient of x in $\frac{3}{8}{x}^{2}-\frac{2}{7}x+\frac{1}{6}$ is $-\frac{2}{7}$.

(v) The constant term in $\frac{\mathrm{\pi }}{2}{x}^{2}+7x-\frac{2}{5}\mathrm{\pi }$ is $-\frac{2}{5}\mathrm{\pi }$.

#### Page No 75:

(i) $\frac{4x-5{x}^{2}+6{x}^{3}}{2x}=\frac{4x}{2x}-\frac{5{x}^{2}}{2x}+\frac{6{x}^{3}}{2x}=2-\frac{5}{2}x+3{x}^{2}$
Here, the highest power of x is 2. So, the degree of the polynomial is 2.

(ii) y2(– y3) = y3 – y5
Here, the highest power of y is 5. So, the degree of the polynomial is 5.

(iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2
Here, the highest power of x is 4. So, the degree of the polynomial is 4.

(iv) $-\frac{1}{2}x+3$
Here, the highest power of x is 1. So, the degree of the polynomial is 1.

(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0.

(vi) x–2(x4 x2) = x2 xx2 + 1
Here, the highest power of x is 2. So, the degree of the polynomial is 2.

#### Page No 75:

(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x5.

(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x8 − 3x.

(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x4 − 3x + 5.

(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.

#### Page No 75:

A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.

(i) $8+x-2{x}^{2}+5{x}^{3}$ is a polynomial in standard form as the powers of x are in ascending order.
(ii) $\frac{2}{3}-3y+4{y}^{2}+2{y}^{3}$ is a polynomial in standard form as the powers of y are in ascending order.
(iii) $2x-3{x}^{2}+6{x}^{3}-{x}^{5}$ is a polynomial in standard form as the powers of x are in ascending order.
(iv) $2+t-{t}^{2}-3{t}^{3}+{t}^{4}$ is a polynomial in standard form as the powers of t are in ascending order.

#### Page No 78:

$=\left(5-0+0\right)\phantom{\rule{0ex}{0ex}}=5$

$=\left(5-12+18\right)\phantom{\rule{0ex}{0ex}}=11$

$=\left(5+8+8\right)\phantom{\rule{0ex}{0ex}}=21$

#### Page No 78:

$=\left(4+0-0+0\right)\phantom{\rule{0ex}{0ex}}=4$

$=\left(4+6-4+40\right)\phantom{\rule{0ex}{0ex}}=46$

$=\left(4-3-1-5\right)\phantom{\rule{0ex}{0ex}}=-5$

#### Page No 78:

$=\left(0-0+6\right)\phantom{\rule{0ex}{0ex}}=6$

$=\left(64-12+6\right)\phantom{\rule{0ex}{0ex}}=58$

$=\left(100+15+6\right)\phantom{\rule{0ex}{0ex}}=121$

#### Page No 78:

$p\left(x\right)={x}^{3}-3{x}^{2}+2x$      .....(1)

Putting x = 0 in (1), we get

$p\left(0\right)={0}^{3}-3×{0}^{2}+2×0=0$

Thus, x = 0 is a zero of p(x).

Putting x = 1 in (1), we get

$p\left(1\right)={1}^{3}-3×{1}^{2}+2×1=1-3+2=0$

Thus, x = 1 is a zero of p(x).

Putting x = 2 in (1), we get

$p\left(2\right)={2}^{3}-3×{2}^{2}+2×2=8-3×4+4=8-12+4=0$

Thus, x = 2 is a zero of p(x).

#### Page No 78:

p(x) = x3 + x2 – 9x – 9         .....(1)

Putting x = 0 in (1), we get

p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0

Thus, x = 0 is not a zero of p(x).

Putting x = 3 in (1), we get

p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).

Putting x = –3 in (1), we get

p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0

Thus, x = –3 is a zero of p(x).

Putting x = –1 in (1), we get

p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0

Thus, x = –1 is a zero of p(x).

#### Page No 78:

= 0
Hence, 4 is the zero of the given polynomial.

Hence, 3 is the zero of the given polynomial.

$=2-2\phantom{\rule{0ex}{0ex}}=0$

Hence, $\frac{2}{5}$ is the zero of the given polynomial.

Hence, $-\frac{1}{2}$ is the zero of the given polynomial.

#### Page No 79:

$=0×\left(-1\right)\phantom{\rule{0ex}{0ex}}=0$
Also,
$p\left(2\right)=\left(2-1\right)\left(2-2\right)$
$=\left(-1\right)×0\phantom{\rule{0ex}{0ex}}=0$

Hence, 1 and 2 are the zeroes of the given polynomial.

$=4-4\phantom{\rule{0ex}{0ex}}=0$
Also,

Hence, 2 and $-3$ are the zeroes of the given polynomial.

Also,

Hence, 0 and 3 are the zeroes of the given polynomial.

#### Page No 79:

It is given that 2 and 0 are the zeroes of the polynomial $f\left(x\right)=2{x}^{3}-5{x}^{2}+ax+b$.
∴ f(2) = 0

Also,
f(0) = 0
$⇒2×{0}^{3}-5×{0}^{2}+a×0+b=0\phantom{\rule{0ex}{0ex}}⇒0-0+0+b=0\phantom{\rule{0ex}{0ex}}⇒b=0$
Putting b = 0 in (1), we get
$2a+0=4\phantom{\rule{0ex}{0ex}}⇒2a=4\phantom{\rule{0ex}{0ex}}⇒a=2$
Thus, the values of a and b are 2 and 0, respectively.

#### Page No 84:

Let f(x) = x4 + 1 and g(x) = x – 1.

Quotient = x3x2x + 1

Remainder = 2

Verification:

Putting x = 1 in f(x), we get

f(1) = 14 + 1 = 1 + 1 = 2 = Remainder, when f(x) = x4 + 1 is divided by g(x) = x – 1

#### Page No 84:

$p\left(x\right)=2{x}^{4}-6{x}^{3}+2{x}^{2}-x+2$ and $g\left(x\right)=x+2$

Quotient = $2{x}^{3}-10{x}^{2}+22x-45$

Remainder = 92

Verification:

Divisor × Quotient + Remainder
$=\left(x+2\right)×\left(2{x}^{3}-10{x}^{2}+22x-45\right)+92\phantom{\rule{0ex}{0ex}}=x\left(2{x}^{3}-10{x}^{2}+22x-45\right)+2\left(2{x}^{3}-10{x}^{2}+22x-45\right)+92\phantom{\rule{0ex}{0ex}}=2{x}^{4}-10{x}^{3}+22{x}^{2}-45x+4{x}^{3}-20{x}^{2}+44x-90+92\phantom{\rule{0ex}{0ex}}=2{x}^{4}-6{x}^{3}+2{x}^{2}-x+2$
= Dividend

Hence verified.

#### Page No 84:

$p\left(x\right)={x}^{3}-6{x}^{2}+9x+3$

â€‹$g\left(x\right)=x-1$

By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).

Putting x = 1 in p(x), we get

$p\left(1\right)={1}^{3}-6×{1}^{2}+9×1+3=1-6+9+3=7$

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

#### Page No 84:

$p\left(x\right)=2{x}^{3}-7{x}^{2}+9x-13$

â€‹$g\left(x\right)=x-3$

By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).

Putting x = 3 in p(x), we get

$p\left(3\right)=2×{3}^{3}-7×{3}^{2}+9×3-13=54-63+27-13=5$

∴ Remainder = 5

Thus, the remainder when p(x) is divided by g(x) is 5.

#### Page No 84:

$p\left(x\right)=3{x}^{4}-6{x}^{2}-8x-2$

â€‹$g\left(x\right)=x-2$

By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).

Putting x = 2 in p(x), we get

$p\left(2\right)=3×{2}^{4}-6×{2}^{2}-8×2-2=48-24-16-2=6$

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

#### Page No 84:

$p\left(x\right)=2{x}^{3}-9{x}^{2}+x+15$

â€‹$g\left(x\right)=2x-3=2\left(x-\frac{3}{2}\right)$

By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = $p\left(\frac{3}{2}\right)$.

Putting $x=\frac{3}{2}$ in p(x), we get

$p\left(\frac{3}{2}\right)=2×{\left(\frac{3}{2}\right)}^{3}-9×{\left(\frac{3}{2}\right)}^{2}+\frac{3}{2}+15=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+15$ $=\frac{27-81+6+60}{4}=\frac{12}{4}=3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

#### Page No 84:

$p\left(x\right)={x}^{3}-2{x}^{2}-8x-1$

â€‹$g\left(x\right)=x+1$

By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).

Putting x = −1 in p(x), we get

$p\left(-1\right)={\left(-1\right)}^{3}-2×{\left(-1\right)}^{2}-8×\left(-1\right)-1=-1-2+8-1=4$

∴ Remainder = 4

Thus, the remainder when p(x) is divided by g(x) is 4.

#### Page No 84:

$p\left(x\right)=2{x}^{3}+{x}^{2}-15x-12$

â€‹$g\left(x\right)=x+2$

By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).

Putting x = −2 in p(x), we get

$p\left(-2\right)=2×{\left(-2\right)}^{3}+{\left(-2\right)}^{2}-15×\left(-2\right)-12=-16+4+30-12=6$

∴ Remainder = 6

Thus, the remainder when p(x) is divided by g(x) is 6.

#### Page No 84:

$p\left(x\right)=6{x}^{3}+13{x}^{2}+3$

â€‹$g\left(x\right)=3x+2=3\left(x+\frac{2}{3}\right)=3\left[x-\left(-\frac{2}{3}\right)\right]$

By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = $p\left(-\frac{2}{3}\right)$.

Putting $x=-\frac{2}{3}$ in p(x), we get

$p\left(-\frac{2}{3}\right)=6×{\left(-\frac{2}{3}\right)}^{3}+13×{\left(-\frac{2}{3}\right)}^{2}+3=-\frac{16}{9}+\frac{52}{9}+3$ $=\frac{-16+52+27}{9}=\frac{63}{9}=7$

∴ Remainder = 7

Thus, the remainder when p(x) is divided by g(x) is 7.

#### Page No 84:

$p\left(x\right)={x}^{3}-6{x}^{2}+2x-4$

â€‹$g\left(x\right)=1-\frac{3}{2}x=-\frac{3}{2}\left(x-\frac{2}{3}\right)$

By remainder theorem, when p(x) is divided by $\left(1-\frac{3}{2}x\right)$, then the remainder = $p\left(\frac{2}{3}\right)$.

Putting $x=\frac{2}{3}$ in p(x), we get

$p\left(\frac{2}{3}\right)={\left(\frac{2}{3}\right)}^{3}-6×{\left(\frac{2}{3}\right)}^{2}+2×\left(\frac{2}{3}\right)-4=\frac{8}{27}-\frac{8}{3}+\frac{4}{3}-4$ $=\frac{8-72+36-108}{27}=-\frac{136}{27}$

∴ Remainder = $-\frac{136}{27}$

Thus, the remainder when p(x) is divided by g(x) is $-\frac{136}{27}$.

#### Page No 84:

$p\left(x\right)=2{x}^{3}+3{x}^{2}-11x-3$

â€‹$g\left(x\right)=\left(x+\frac{1}{2}\right)=\left[x-\left(-\frac{1}{2}\right)\right]$

By remainder theorem, when p(x) is divided by $\left(x+\frac{1}{2}\right)$, then the remainder = $p\left(-\frac{1}{2}\right)$.

Putting $x=-\frac{1}{2}$ in p(x), we get

$p\left(-\frac{1}{2}\right)=2×{\left(-\frac{1}{2}\right)}^{3}+3×{\left(-\frac{1}{2}\right)}^{2}-11×\left(-\frac{1}{2}\right)-3=-\frac{1}{4}+\frac{3}{4}+\frac{11}{2}-3$ $=\frac{-1+3+22-12}{4}=\frac{12}{4}=3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

#### Page No 84:

$p\left(x\right)={x}^{3}-a{x}^{2}+6x-a$

â€‹$g\left(x\right)=x-a$

By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).

Putting x = a in p(x), we get

$p\left(a\right)={a}^{3}-a×{a}^{2}+6×a-a={a}^{3}-{a}^{3}+6a-a=5a$

∴ Remainder = 5a

Thus, the remainder when p(x) is divided by g(x) is 5a.

#### Page No 84:

Let $f\left(x\right)=2{x}^{3}+{x}^{2}-ax+2$ and $g\left(x\right)=2{x}^{3}-3{x}^{2}-3x+a$.

By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).

Putting x = 2 in f(x), we get

$f\left(2\right)=2×{2}^{3}+{2}^{2}-a×2+2=16+4-2a+2=-2a+22$

By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).

Putting x = 2 in g(x), we get

$g\left(2\right)=2×{2}^{3}-3×{2}^{2}-3×2+a=16-12-6+a=-2+a$

It is given that,

$f\left(2\right)=g\left(2\right)\phantom{\rule{0ex}{0ex}}⇒-2a+22=-2+a\phantom{\rule{0ex}{0ex}}⇒-3a=-24\phantom{\rule{0ex}{0ex}}⇒a=8$
Thus, the value of a is 8.

#### Page No 84:

$\mathrm{Let}:\phantom{\rule{0ex}{0ex}}p\left(x\right)={x}^{4}-2{x}^{3}+3{x}^{2}-ax+b$

Thus, we have:

And,

Now,

$⇒2b=16\phantom{\rule{0ex}{0ex}}⇒b=8$
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,

Also,

Thus, we have:

#### Page No 85:

$p\left(x\right)={x}^{3}-5{x}^{2}+4x-3$

$g\left(x\right)=x-2$

Putting x = 2 in p(x), we get

$p\left(2\right)={2}^{3}-5×{2}^{2}+4×2-3=8-20+8-3=-7\ne 0$

Therefore, by factor theorem, (x − 2) is not a factor of p(x).

Hence, p(x) is not a multiple of g(x).

#### Page No 85:

$p\left(x\right)=2{x}^{3}-11{x}^{2}-4x+5$

$g\left(x\right)=2x+1=2\left(x+\frac{1}{2}\right)=2\left[x-\left(-\frac{1}{2}\right)\right]$

Putting $x=-\frac{1}{2}$ in p(x), we get

$p\left(-\frac{1}{2}\right)=2×{\left(-\frac{1}{2}\right)}^{3}-11×{\left(-\frac{1}{2}\right)}^{2}-4×\left(-\frac{1}{2}\right)+5=-\frac{1}{4}-\frac{11}{4}+2+5$ $=-\frac{12}{4}+7=-3+7=4\ne 0$

Therefore, by factor theorem, (2x + 1) is not a factor of p(x).

Hence, g(x) is not a factor of p(x).

#### Page No 90:

Let:
p(x) = x3 – 8
Now,
$g\left(x\right)=0⇒x-2=0⇒x=2$
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
$p\left(2\right)=\left({2}^{3}-8\right)=0$
Hence, (x $-$ 2) is a factor of the given polynomial.

#### Page No 90:

Let:
p(x) = 2x3 + 7x2 – 24x – 45
Now,
$x-3=0⇒x=3$
By the factor theorem, (x $-$ 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:

Hence, (x $-$ 3) is a factor of the given polynomial.

#### Page No 90:

Let:
p(x) = 2x4 + 9x3 + 6x2 – 11x – 6
Here,
$x-1=0⇒x=1$â€‹
By the factor theorem, (x $-$ 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:

Hence, (x $-$ 1) is a factor of the given polynomial.

#### Page No 91:

Let:
p(x) = x4x2 – 12
Here,

By the factor theorem, (x + 2) is a factor of the given polynomial if p ($-$2) = 0.
Thus, we have:

Hence, (x + 2) is a factor of the given polynomial.

#### Page No 91:

$p\left(x\right)=69+11x-{x}^{2}+{x}^{3}$

$g\left(x\right)=x+3$

Putting x = −3 in p(x), we get

$p\left(-3\right)=69+11×\left(-3\right)-{\left(-3\right)}^{2}+{\left(-3\right)}^{3}=69-33-9-27=0$

Therefore, by factor theorem, (x + 3) is a factor of p(x).

Hence, g(x) is a factor of p(x).

#### Page No 91:

Let:
$p\left(x\right)=2{x}^{3}+9{x}^{2}-11x-30$
Here,
$x+5=0⇒x=-5$
By the factor theorem, (x + 5) is a factor of the given polynomial if p ($-$5) = 0.
Thus, we have:

Hence, (x + 5) is a factor of the given polynomial.

#### Page No 91:

Let:
$p\left(x\right)=2{x}^{4}+{x}^{3}-8{x}^{2}-x+6$
Here,
$2x-3=0⇒x=\frac{3}{2}$
By the factor theorem, (2x $-$ 3) is a factor of the given polynomial if $p\left(\frac{3}{2}\right)=0$.
Thus, we have:

Hence, (2x $-$ 3) is a factor of the given polynomial.

#### Page No 91:

$p\left(x\right)=3{x}^{3}+{x}^{2}-20x+12$

$g\left(x\right)=3x-2=3\left(x-\frac{2}{3}\right)$

Putting $x=\frac{2}{3}$ in p(x), we get

$p\left(\frac{2}{3}\right)=3×{\left(\frac{2}{3}\right)}^{3}+{\left(\frac{2}{3}\right)}^{2}-20×\frac{2}{3}+12=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12$ $=\frac{8+4-120+108}{9}=\frac{120-120}{9}=0$

Therefore, by factor theorem, (3x − 2) is a factor of p(x).

Hence, g(x) is a factor of p(x).

#### Page No 91:

Let:
$p\left(x\right)=7{x}^{2}-4\sqrt{2}x-6$
Here,
$x-\sqrt{2}=0⇒x=\sqrt{2}$
By the factor theorem, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial if $p\left(\sqrt{2}\right)=0$
Thus, we have:

Hence, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial.

#### Page No 91:

Let:
$p\left(x\right)=2\sqrt{2}{x}^{2}+5x+\sqrt{2}$
Here,
$x+\sqrt{2}=0⇒x=-\sqrt{2}$
By the factor theorem, $\left(x+\sqrt{2}\right)$ will be a factor of the given polynomial if $p\left(-\sqrt{2}\right)$ = 0.
Thus, we have:

Hence, $\left(x+\sqrt{2}\right)$ is a factor of the given polynomial.

#### Page No 91:

Let f(p) = p10 – 1 and g(p) = p11 – 1.

Putting p = 1 in f(p), we get

f(1) = 110 − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).

Now, putting p = 1 in g(p), we get

g(1) = 111 − 1 = 1 − 1 = 0

Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).

#### Page No 91:

Let:
$f\left(x\right)=2{x}^{3}+9{x}^{2}+x+k$

Hence, the required value of k is $-$12.

#### Page No 91:

Let:
$f\left(x\right)=2{x}^{3}-3{x}^{2}-18x+a$

Hence, the required value of a is $-$8.

#### Page No 91:

Let f(x) = ax3 x2 – 2x + 4a – 9

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

Using factor theorem,  we have

f(−1) = 0
$⇒a×{\left(-1\right)}^{3}+{\left(-1\right)}^{2}-2×\left(-1\right)+4a-9=0\phantom{\rule{0ex}{0ex}}⇒-a+1+2+4a-9=0\phantom{\rule{0ex}{0ex}}⇒3a-6=0\phantom{\rule{0ex}{0ex}}⇒3a=6\phantom{\rule{0ex}{0ex}}⇒a=2$
Thus, the value of a is 2.

#### Page No 91:

Let f(x) = x5 – 4a2 x3 + 2x + 2a +3

It is given that (+ 2a) is a factor of f(x).

Using factor theorem,  we have

f(−2a) = 0
$⇒{\left(-2a\right)}^{5}-4{a}^{2}×{\left(-2a\right)}^{3}+2×\left(-2a\right)+2a+3=0\phantom{\rule{0ex}{0ex}}⇒-32{a}^{5}-4{a}^{2}×\left(-8{a}^{3}\right)+2×\left(-2a\right)+2a+3=0\phantom{\rule{0ex}{0ex}}⇒-32{a}^{5}+32{a}^{5}-4a+2a+3=0\phantom{\rule{0ex}{0ex}}⇒-2a+3=0$
$⇒2a=3\phantom{\rule{0ex}{0ex}}⇒a=\frac{3}{2}$
Thus, the value of a is $\frac{3}{2}$.

#### Page No 91:

Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m

It is given that $\left(2x-1\right)=2\left(x-\frac{1}{2}\right)$ is a factor of f(x).

Using factor theorem,  we have

$f\left(\frac{1}{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒8×{\left(\frac{1}{2}\right)}^{4}+4×{\left(\frac{1}{2}\right)}^{3}-16×{\left(\frac{1}{2}\right)}^{2}+10×\frac{1}{2}+m=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}+\frac{1}{2}-4+5+m=0$
$⇒2+m=0\phantom{\rule{0ex}{0ex}}⇒m=-2$
Thus, the value of m is −2.

#### Page No 91:

Let:
$f\left(x\right)={x}^{4}-{x}^{3}-11{x}^{2}-x+a$
Now,
$x+3=0⇒x=-3$
By the factor theorem, .
Thus, we have:

Also,

Hence,

#### Page No 91:

Let:
$f\left(x\right)={x}^{3}-3{x}^{2}-13x+15$
And,
$g\left(x\right)={x}^{2}+2x-3$
$={x}^{2}+x-3x-3\phantom{\rule{0ex}{0ex}}=x\left(x-1\right)+3\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x+3\right)$
Now, .
For this, we must have:

Thus, we have:

And,

. So, .
Hence, .

#### Page No 91:

Let:
$f\left(x\right)={x}^{3}+a{x}^{2}+bx+6$

Now,
$x-3=0⇒x=3$
By the factor theorem, we can say:

Now,

Thus, we have:

Subtracting (1) from (2), we get:
a = $-$3
By putting the value of a in (1), we get the value of b, i.e., $-$1.
a = $-$3 and b = $-$1

#### Page No 91:

Let:
$f\left(x\right)={x}^{3}-10{x}^{2}+ax+b$
Now,
$x-1=0⇒x=1$
By the factor theorem, we can say:
.
Thus, we have:

∴
Also,
$x-2=0⇒x=2$
By the factor theorem, we can say:
.
Thus, we have:

Putting the value of a, we get the value of b, i.e., $-$14.
∴ a = 23 and b = $-$14

#### Page No 91:

Let:
$f\left(x\right)={x}^{4}+a{x}^{3}-7{x}^{2}-8x+b$
Now,
$x+2=0⇒x=-2$
By the factor theorem, we can say:
.
Thus, we have:

∴
Also,
$x+3=0⇒x=-3$
By the factor theorem, we can say:
.
Thus, we have:

∴

Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

#### Page No 91:

Let f(x) = px2 + 5x + r

It is given that (x – 2) is a factor of f(x).

Using factor theorem,  we have

Also,  $\left(x-\frac{1}{2}\right)$ is a factor of f(x).

Using factor theorem,  we have

From (1) and (2), we have

$4p+r=p+4r\phantom{\rule{0ex}{0ex}}⇒4p-p=4r-r\phantom{\rule{0ex}{0ex}}⇒3p=3r\phantom{\rule{0ex}{0ex}}⇒p=r$

#### Page No 91:

Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.

${x}^{2}-3x+2\phantom{\rule{0ex}{0ex}}={x}^{2}-2x-x+2\phantom{\rule{0ex}{0ex}}=x\left(x-2\right)-1\left(x-2\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-2\right)$
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).

Putting x = 1 in f(x), we get

f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).

Putting x = 2 in f(x), we get

f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0

By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).

Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).

Hence, f(x) = 2x4 – 5x3 + 2x2 – x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.

#### Page No 91:

Let k be added to 2x4 – 5x3 + 2x2 – – 3 so that the result is exactly divisible by (– 2). Here, k is a constant.

∴ f(x) = 2x4 – 5x3 + 2x2 – – 3 + k is exactly divisible by (– 2).

Using factor theorem, we have

$f\left(2\right)=0\phantom{\rule{0ex}{0ex}}⇒2×{2}^{4}-5×{2}^{3}+2×{2}^{2}-2-3+k=0\phantom{\rule{0ex}{0ex}}⇒32-40+8-5+k=0\phantom{\rule{0ex}{0ex}}⇒-5+k=0\phantom{\rule{0ex}{0ex}}⇒k=5$
Thus, 5 must be added to 2x4 – 5x3 + 2x2 – – 3 so that the result is exactly divisible by (– 2).

#### Page No 91:

Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x+ 2x – 3) using long division method, we have

Here, the remainder obtained is (2x + 9).

Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x+ 2x – 3).

#### Page No 91:

Let f(x) = xn an

Putting x = −a in f(x), we get

f(−a) = (−a)n an

If n is any odd positive integer, then

f(−a) = (−a)n an = −an an = 0

Therefore, by factor theorem, (x + a) is a factor of (xn an) for any odd positive integer.

#### Page No 92:

(c)

Clearly, $\sqrt{2}{x}^{2}-\sqrt{3}x+6$ is a polynomial in one variable because it has only non-negative integral powers of x.

#### Page No 92:

(d) ${x}^{2}+\frac{2{x}^{3/2}}{\sqrt{x}}+6$

We have:

${x}^{2}+\frac{2{x}^{\frac{3}{2}}}{\sqrt{x}}+6={x}^{2}+2{x}^{\frac{3}{2}}{x}^{-\frac{1}{2}}+6$
$={x}^{2}+2x+6$

It a polynomial because it has only non-negative integral powers of x.

#### Page No 93:

(c) y

y is a polynomial because it has a non-negative integral power 1.

#### Page No 93:

(d) −4

$-$4 is a constant polynomial of degree zero.

#### Page No 93:

(d) 0

0 is a polynomial whose degree is not defined.

#### Page No 93:

(d) x2 + 5x + 4

${x}^{2}+5x+4$ is a polynomial of degree 2. So, it is a quadratic polynomial.

#### Page No 93:

(b) x + 1

Clearly, $x+1$ is a polynomial of degree 1. So, it is a linear polynomial.

#### Page No 93:

(b) x2 + 4

Clearly, ${x}^{2}+4$ is an expression having two non-zero terms. So, it is a binomial.

#### Page No 93:

(d) 0

$\sqrt{3}$ is a constant term, so it is a polynomial of degree 0.

#### Page No 93:

(c) not defined

Degree of the zero polynomial is not defined.

#### Page No 93:

(d) not defined

Zero of the zero polynomial is not defined.

#### Page No 93:

(d) 8

Let:
$p\left(x\right)=\left(x+4\right)$

Thus, we have:
$p\left(x\right)+p\left(-x\right)=\left\{\left(x+4\right)+\left(-x+4\right)\right\}$
= 4 + 4
=8

#### Page No 93:

(b) 1

= 8 $-$ 8 + 1
= 1

#### Page No 93:

p(x) = 5x – 4x2 + 3

Putting x = –1 in p(x), we get

p(–1) = 5 × (–1) – 4 × (–1)2 + 3 = –5 – 4 + 3 = –6

Hence, the correct answer is option (d).

#### Page No 93:

Let f(x) = x51 + 51

By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).

Putting x = −1 in f(x), we get

f(−1) = (−1)51 + 51 = −1 + 51 = 50

∴ Remainder = 50

Thus, the remainder when (x51 + 51) is divided by (x + 1) is 50.

Hence, the correct answer is option (d).

(c) 2

#### Page No 93:

(d) 21

$x-2=0⇒x=2$
By the remainder theorem, we know that when p(x) is divided by (x $-$ 2), the remainder is p(2).
Thus, we have:

#### Page No 94:

(d) 4

$x+2=0⇒x=-2$
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p($-$2).
Now, we have:

#### Page No 94:

(c) −2

$2x-1=0⇒x=\frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x $-$ 1), the remainder is $p\left(\frac{1}{2}\right)$.
Now, we have:

#### Page No 94:

By remainder theorem, when p(x) = x3 – ax2 + x is divided by (x – a), then the remainder = p(a).

Putting xa in p(x), we get

p(a) = a3 – a × a2 + aa3 – a3 + aa

∴ Remainder = a

Hence, the correct answer is option (b).

#### Page No 94:

(c) −a

$x+a=0⇒x=-a$
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:

#### Page No 94:

(b) x3 − 2x2 − x − 2

Let:
f(x) = x3 − 2x2 − x − 2
By the factor theorem, (x + 1) will be a factor of f(x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is a factor of $f\left(x\right)={x}^{3}+2{x}^{2}-x-2$.

#### Page No 94:

The zero of the polynomial p(x) can be obtained by putting p(x) = 0.

$p\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2x+5=0\phantom{\rule{0ex}{0ex}}⇒2x=-5\phantom{\rule{0ex}{0ex}}⇒x=-\frac{5}{2}$
Hence, the correct answer is option (b).

#### Page No 94:

The given polynomial is p(x) = x2 + x – 6.

Putting x = 2 in p(x), we get

p(2) = 22 + 2 – 6 = 4 + 2 – 6 = 0

Therefore, x = 2 is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

p(–3) = (–3)2 – 3 – 6 = 9 – 9 = 0

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, 2 and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

#### Page No 94:

The given polynomial is p(x) = 2x2 + 5x – 3.

Putting $x=\frac{1}{2}$ in p(x), we get

$p\left(\frac{1}{2}\right)=2×{\left(\frac{1}{2}\right)}^{2}+5×\frac{1}{2}-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –3 in p(x), we get

$p\left(-3\right)=2×{\left(-3\right)}^{2}+5×\left(-3\right)-3=18-15-3=0$

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –3 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (b).

#### Page No 94:

The given polynomial is p(x) = 2x2 + 7x – 4.

Putting $x=\frac{1}{2}$ in p(x), we get

$p\left(\frac{1}{2}\right)=2×{\left(\frac{1}{2}\right)}^{2}+7×\frac{1}{2}-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial p(x).

Putting x = –4 in p(x), we get

$p\left(-4\right)=2×{\left(-4\right)}^{2}+7×\left(-4\right)-4=32-28-4=32-32=0$

Therefore, x = –4 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and –4 are the zeroes of the given polynomial p(x).

Hence, the correct answer is option (c).

(b) 5

#### Page No 94:

(b) m = 7, n = −18

Let:
$p\left(x\right)={x}^{3}+10{x}^{2}+mx+n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0

Now,
$x-1=0⇒x=1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0

By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

#### Page No 94:

(a) 1

Let:
$p\left(x\right)={x}^{100}+2{x}^{99}+k$
Now,
$x+1=0⇒x=-1$

#### Page No 94:

(d) −3

Let:
$p\left(x\right)=2{x}^{3}-k{x}^{2}+3x+10\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$

#### Page No 94:

(b) 0, 3

Let:
$p\left(x\right)={x}^{2}-3x$
Now, we have:
$p\left(x\right)=0⇒{x}^{2}-3x=0$

#### Page No 95:

$p\left(x\right)=3{x}^{2}-1$