RS AGGARWAL CLASS 9 Chapter 2 POLYNOMIALS EXERCISE 2D

 EXERCISE 2D

PAGE NO-90

Question 1:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x³ – 8, g(x) = x – 2

Answer 1:

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

Question 2:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x³ + 7x² – 24x – 45, g(x) = x – 3

Answer 2:

Let:
p(x) = 2x³ + 7x² – 24x – 45
Now,
$x-3=0\Rightarrow x=3$$x-3=0\Rightarrow x=3$
By the factor theorem, (x $-$$-$ 3) is a factor of the given polynomial if p(3) = 0.
Thus, we have:
$p\left(3\right)=\left(2\times 3^3-7\times 3^2-24\times 3-45\right)=\left(54+63-72-45\right)=0$$$\begin{gathered} p\left(3\right)=\left(2\times 3^3-7\times 3^2-24\times 3-45\right) \\ =\left(54+63-72-45\right) \\ =0 \end{gathered}$$
Hence, (x $-$$-$ 3) is a factor of the given polynomial.

Question 3:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x⁴ + 9x³ + 6x² – 11x – 6, g(x) = x – 1

Answer 3:

Let:
p(x) = 2x⁴ + 9x³ + 6x² – 11x – 6
Here, 
$x-1=0\Rightarrow x=1$$x-1=0\Rightarrow x=1$​
By the factor theorem, (x $-$$-$ 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
$p\left(1\right)=\left(2\times 1^4+9\times 1^3+6\times 1^2-11\times 1-6\right)=\left(2+9+6-11-6\right)=0$$$\begin{gathered} p\left(1\right)=\left(2\times 1^4+9\times 1^3+6\times 1^2-11\times 1-6\right) \\ =\left(2+9+6-11-6\right) \\ =0 \end{gathered}$$
Hence, (x $-$$-$ 1) is a factor of the given polynomial.

PAGE NO-91

Question 4:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x⁴ – x² – 12, g(x) = x + 2

Answer 4:

Let:
p(x) = x⁴ – x² – 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.

Question 5:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 69 + 11x – x² + x³, g(x) = x + 3

Answer 5:

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

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

Putting x = −3 in p(x), we get

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

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

Hence, g(x) is a factor of p(x).

Question 6:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x³ + 9x² – 11x – 30, g(x) = x + 5

Answer 6:

Let:

Here, 

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.

Question 7:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 2x⁴ + x³ – 8x² – x + 6, g(x) = 2x – 3

Answer 7:

Let:

Here, 

By the factor theorem, (2x $-$ 3) is a factor of the given polynomial if .
Thus, we have:


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

Question 8:

Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2

Answer 8:

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

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

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

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

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

Hence, g(x) is a factor of p(x).

Question 9:

Using factor theorem, show that g(x) is a factor of p(x), when
$p\left(x\right)=7x^2-4\sqrt{2}x-6, g\left(x\right)=x-\sqrt{2}$

Answer 9:

Let:

Here, 

By the factor theorem, is a factor of the given polynomial if 
Thus, we have:

Hence,  is a factor of the given polynomial.

Question 10:

Using factor theorem, show that g(x) is a factor of p(x), when
$p\left(x\right)=2\sqrt{2}{x}^2+5x+\sqrt{2}, g\left(x\right)=x+\sqrt{2}$

Answer 10:

Let:

Here,

By the factor theorem,  will be a factor of the given polynomial if  = 0.
Thus, we have:

Hence,  is a factor of the given polynomial.

Question 11:

Show that (p – 1) is a factor of (p¹⁰ – 1) and also of (p¹¹ – 1).

Answer 11:

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).

Question 12:

Find the value of k for which (x − 1) is a factor of (2x³ + 9x² + x + k).

Answer 12:

 Let:


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

Question 13:

Find the value of a for which (x − 4) is a factor of (2x³ − 3x² − 18x + a).

Answer 13:

Let:


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

Question 14:

Find the value of a for which (x + 1) is a factor of (ax³ + x² – 2x + 4a – 9).

Answer 14:

Let f(x) = ax³ + x² – 2x + 4a – 9

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

Using factor theorem,  we have

f(−1) = 0
$$\begin{gathered} \Rightarrow a\times {\left(-1\right)}^3+{\left(-1\right)}^2-2\times \left(-1\right)+4a-9=0 \\ \Rightarrow -a+1+2+4a-9=0 \\ \Rightarrow 3a-6=0 \\ \Rightarrow 3a=6 \\ \Rightarrow a=2 \end{gathered}$$
Thus, the value of a is 2.

Question 15:

Find the value of a for which (x + 2a) is a factor of (x⁵ – 4a² x³ + 2x + 2a +3).

Answer 15:

Let f(x) = x⁵ – 4a² x³ + 2x + 2a +3

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

Using factor theorem,  we have

f(−2a) = 0
$$\begin{gathered} \Rightarrow {\left(-2a\right)}^5-4a^2\times {\left(-2a\right)}^3+2\times \left(-2a\right)+2a+3=0 \\ \Rightarrow -32a^5-4a^2\times \left(-8a^3\right)+2\times \left(-2a\right)+2a+3=0 \\ \Rightarrow -32a^5+32a^5-4a+2a+3=0 \\ \Rightarrow -2a+3=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2a=3 \\ \Rightarrow a=\frac{3}{2} \end{gathered}$$
Thus, the value of a is $\frac{3}{2}$.

Question 16:

Find the value of m for which (2x – 1) is a factor of (8x⁴ + 4x³ – 16x² + 10x + m).

Answer 16:

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

$$\begin{gathered} f\left(\frac{1}{2}\right)=0 \\ \Rightarrow 8\times {\left(\frac{1}{2}\right)}^4+4\times {\left(\frac{1}{2}\right)}^3-16\times {\left(\frac{1}{2}\right)}^2+10\times \frac{1}{2}+m=0 \\ \Rightarrow \frac{1}{2}+\frac{1}{2}-4+5+m=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2+m=0 \\ \Rightarrow m=-2 \end{gathered}$$
Thus, the value of m is −2.

Question 17:

Find the value of a for which the polynomial (x⁴ − x³ − 11x² − x + a) is divisible by (x + 3).

Answer 17:

Let:

Now, 

By the factor theorem, .
Thus, we have:

Also,

Hence, 

Question 18:

Without actual division, show that (x³ − 3x² − 13x + 15) is exactly divisible by (x² + 2x − 3).

Answer 18:

Let:

And,

    
Now, .
For this, we must have:

Thus, we have:

And,

. So, .
Hence, .

Question 19:

If (x³ + ax² + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

Answer 19:

Let:


Now, 

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

Question 20:

Find the values of a and b so that the polynomial (x³ − 10x² + ax + b) is exactly divisible by (x −1) as well as (x − 2).

Answer 20:

Let:

Now,

By the factor theorem, we can say:
.
Thus, we have:

∴ 
Also,

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

Question 21:

Find the values of a and b so that the polynomial (x⁴ + ax³ − 7x² − 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Answer 21:

Let:

Now,

By the factor theorem, we can say:
.
Thus, we have:

∴ 
Also,

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

Question 22:

If both (x – 2) and $\left(x-\frac{1}{2}\right)$ are factors of px² + 5x + r, prove that p = r.

Answer 22:

Let f(x) = px² + 5x + r

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

Using factor theorem,  we have

$$\begin{gathered} f\left(2\right)=0 \\ \Rightarrow p\times 2^2+5\times 2+r=0 \\ \Rightarrow 4p+r=-10 .....\left(1\right) \end{gathered}$$
Also,  $\left(x-\frac{1}{2}\right)$ is a factor of f(x).

Using factor theorem,  we have

$$\begin{gathered} f\left(\frac{1}{2}\right)=0 \\ \Rightarrow p\times {\left(\frac{1}{2}\right)}^2+5\times \frac{1}{2}+r=0 \\ \Rightarrow \frac{p}{4}+r=-\frac{5}{2} \\ \Rightarrow p+4r=-10 .....\left(2\right) \end{gathered}$$
From (1) and (2), we have

$$\begin{gathered} 4p+r=p+4r \\ \Rightarrow 4p-p=4r-r \\ \Rightarrow 3p=3r \\ \Rightarrow p=r \end{gathered}$$

 

Question 23:

Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

Answer 23:

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

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

Question 24:

What must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2)?

Answer 24:

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

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

Using factor theorem, we have

$$\begin{gathered} f\left(2\right)=0 \\ \Rightarrow 2\times 2^4-5\times 2^3+2\times 2^2-2-3+k=0 \\ \Rightarrow 32-40+8-5+k=0 \\ \Rightarrow -5+k=0 \\ \Rightarrow k=5 \end{gathered}$$
Thus, 5 must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2).

Question 25:

What must be subtracted from (x⁴ + 2x³ – 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x – 3)?

Answer 25:

Dividing (x⁴ + 2x³ – 2x² + 4x + 6) by (x² + 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 (x² + 2x – 3).

Question 26:

Use factor theorem to prove that (x + a) is a factor of (xⁿ + aⁿ) for any odd positive integer.

Answer 26:

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.