myhelper: 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,
$g (x) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$ $g (x) = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, (x – 2) is a factor of the given polynomial if p(2) = 0.
Thus, we have:
$p (2) = (2^{3} - 8) = 0$ $p (2) = (2^{3} - 8) = 0$
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:
$p (3) = (2 \times 3^{3} - 7 \times 3^{2} - 24 \times 3 - 45) = (54 + 63 - 72 - 45) = 0$ $p (3) = (2 \times 3^{3} - 7 \times 3^{2} - 24 \times 3 - 45) = (54 + 63 - 72 - 45) = 0$
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:
$p (1) = (2 \times 1^{4} + 9 \times 1^{3} + 6 \times 1^{2} - 11 \times 1 - 6) = (2 + 9 + 6 - 11 - 6) = 0$ $p (1) = (2 \times 1^{4} + 9 \times 1^{3} + 6 \times 1^{2} - 11 \times 1 - 6) = (2 + 9 + 6 - 11 - 6) = 0$
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,
$x + 2 = 0 \Rightarrow x = - 2$
By the factor theorem, (x + 2) is a factor of the given polynomial if p ( $-$ 2) = 0.
Thus, we have:
$p (- 2) = [{(- 2)}^{4} - {(- 2)}^{2} - 12] = (16 - 4 - 12) = 0$
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:

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

$g (x) = x + 3$

Putting x = −3 in p(x), we get

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

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:
$p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30$
Here,
$x + 5 = 0 \Rightarrow x = - 5$
By the factor theorem, (x + 5) is a factor of the given polynomial if p ( $-$ 5) = 0.
Thus, we have:
$p (- 5) = [2 \times {(- 5)}^{3} + 9 \times {(- 5)}^{2} - 11 \times (- 5) - 30] = (- 250 + 225 + 55 - 30) = 0$
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:
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6$
Here,
$2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$
By the factor theorem, (2x $-$ 3) is a factor of the given polynomial if $p (\frac{3}{2}) = 0$ .
Thus, we have:

$p (\frac{3}{2}) = [2 \times {(\frac{3}{2})}^{4} + {(\frac{3}{2})}^{3} - 8 \times {(\frac{3}{2})}^{2} - (\frac{3}{2}) + 6] = (\frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} + 6) = 0$
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:

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

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

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

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

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
$p (x) = 7 x^{2} - 4 \sqrt{2} x - 6, g (x) = x - \sqrt{2}$

Answer 9:

Let:
$p (x) = 7 x^{2} - 4 \sqrt{2} x - 6$
Here,
$x - \sqrt{2} = 0 \Rightarrow x = \sqrt{2}$
By the factor theorem, $(x - \sqrt{2})$ is a factor of the given polynomial if $p (\sqrt{2}) = 0$
Thus, we have:
$p (\sqrt{2}) = [7 \times {(\sqrt{2})}^{2} - 4 \sqrt{2} \times \sqrt{2} - 6] = (14 - 8 - 6) = 0$
Hence, $(x - \sqrt{2})$ is a factor of the given polynomial.

Question 10:

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

Answer 10:

Let:
$p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}$
Here,
$x + \sqrt{2} = 0 \Rightarrow x = - \sqrt{2}$
By the factor theorem, $(x + \sqrt{2})$ will be a factor of the given polynomial if $p (- \sqrt{2})$ = 0.
Thus, we have:
$p (- \sqrt{2}) = [2 \sqrt{2} \times {(- \sqrt{2})}^{2} + 5 \times (- \sqrt{2}) + \sqrt{2}] = (4 \sqrt{2} - 5 \sqrt{2} + \sqrt{2}) = 0$
Hence, $(x + \sqrt{2})$ 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:
$f (x) = 2 x^{3} + 9 x^{2} + x + k$
$(x - 1) is a factor of f (x) = 2 x^{3} + 9 x^{2} + x + k . \Rightarrow f (1) = 0 \Rightarrow 2 \times 1^{3} + 9 \times 1^{2} + 1 + k = 0 \Rightarrow 12 + k = 0 \Rightarrow k = - 12$
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:
$f (x) = 2 x^{3} - 3 x^{2} - 18 x + a$
$(x - 4) is a factor of f (x) = 2 x^{3} - 3 x^{2} - 18 x + a . \Rightarrow f (4) = 0 \Rightarrow 2 \times 4^{3} - 3 \times 4^{2} - 18 \times 4 + a = 0 \Rightarrow 8 + a = 0 \Rightarrow a = - 8$
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
$\Rightarrow a \times {(- 1)}^{3} + {(- 1)}^{2} - 2 \times (- 1) + 4 a - 9 = 0 \Rightarrow - a + 1 + 2 + 4 a - 9 = 0 \Rightarrow 3 a - 6 = 0 \Rightarrow 3 a = 6 \Rightarrow a = 2$
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
$\Rightarrow {(- 2 a)}^{5} - 4 a^{2} \times {(- 2 a)}^{3} + 2 \times (- 2 a) + 2 a + 3 = 0 \Rightarrow - 32 a^{5} - 4 a^{2} \times (- 8 a^{3}) + 2 \times (- 2 a) + 2 a + 3 = 0 \Rightarrow - 32 a^{5} + 32 a^{5} - 4 a + 2 a + 3 = 0 \Rightarrow - 2 a + 3 = 0$
$\Rightarrow 2 a = 3 \Rightarrow a = \frac{3}{2}$
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 $(2 x - 1) = 2 (x - \frac{1}{2})$ is a factor of f(x).

Using factor theorem, we have

$f (\frac{1}{2}) = 0 \Rightarrow 8 \times {(\frac{1}{2})}^{4} + 4 \times {(\frac{1}{2})}^{3} - 16 \times {(\frac{1}{2})}^{2} + 10 \times \frac{1}{2} + m = 0 \Rightarrow \frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$
$\Rightarrow 2 + m = 0 \Rightarrow m = - 2$
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:
$f (x) = x^{4} - x^{3} - 11 x^{2} - x + a$
Now,
$x + 3 = 0 \Rightarrow x = - 3$
By the factor theorem, $f (x) is exactly divisible by (x + 3) if f (- 3) = 0$ .
Thus, we have:
$f (- 3) = [{(- 3)}^{4} - {(- 3)}^{3} - 11 \times {(- 3)}^{2} - (- 3) + a] = (81 + 27 - 99 + 3 + a) = 12 + a$
Also,
$f (- 3) = 0 \Rightarrow 12 + a = 0 \Rightarrow a = - 12$
Hence, $f (x) is exactly divisible by (x + 3) when a is - 12 .$

Question 18:

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

Answer 18:

Let:
$f (x) = x^{3} - 3 x^{2} - 13 x + 15$
And,
$g (x) = x^{2} + 2 x - 3$
$= x^{2} + x - 3 x - 3 = x (x - 1) + 3 (x - 1) = (x - 1) (x + 3)$
Now, $f (x) will be exactly divisible by g (x) if it is exactly divisible by (x - 1) as well as (x + 3)$ .
For this, we must have:
$f (1) = 0 and f (- 3) = 0$
Thus, we have:
$f (1) = (1^{3} - 3 \times 1^{2} - 13 \times 1 + 15) = (1 - 3 - 13 + 15) = 0$
And,
$f (- 3) = [{(- 3)}^{3} - 3 \times {(- 3)}^{2} - 13 \times (- 3) + 15] = (- 27 - 27 + 39 + 15) = 0$
$f (x) is exactly divisible by (x - 1) as well as (x + 3)$ . So, $f (x) is exactly divisible by (x - 1) (x + 3)$ .
Hence, $f (x) is exactly divisible b y x^{2} + 2 x - 3$ .

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:
$f (x) = x^{3} + a x^{2} + b x + 6$
$(x - 2) is a factor of f (x) = x^{3} + a x^{2} + b x + 6 . \Rightarrow f (2) = 0 \Rightarrow 2^{3} + a \times 2^{2} + b \times 2 + 6 = 0 \Rightarrow 14 + 4 a + 2 b = 0 \Rightarrow 4 a + 2 b = - 14 \Rightarrow 2 a + b = - 7 . . . (1)$
Now,
$x - 3 = 0 \Rightarrow x = 3$
By the factor theorem, we can say:
$When f (x) will be divided by (x - 3), 3 will be its remainder . \Rightarrow f (3) = 3$

Now,
$f (3) = 3^{3} + a \times 3^{2} + b \times 3 + 6 = (27 + 9 a + 3 b + 6) = 33 + 9 a + 3 b$
Thus, we have:
$f (3) = 3 \Rightarrow 33 + 9 a + 3 b = 3 \Rightarrow 9 a + 3 b = - 30 \Rightarrow 3 a + b = - 10 . . . (2)$
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:
$f (x) = x^{3} - 10 x^{2} + a x + b$
Now,
$x - 1 = 0 \Rightarrow x = 1$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x - 1) if f (1) = 0$ .
Thus, we have:
$f (1) = 1^{3} - 10 \times 1^{2} + a \times 1 + b = (1 - 10 + a + b) = - 9 + a + b$
∴ $f (1) = 0 \Rightarrow a + b = 9 . . . (1)$
Also,
$x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x - 2) if f (2) = 0$ .
Thus, we have:
$f (2) = 2^{3} - 10 \times 2^{2} + a \times 2 + b = (8 - 40 + 2 a + b) = - 32 + 2 a + b$
∴ $f (2) = 0 \Rightarrow 2 a + b = 32 . . . (2)$

$Subtracting (1) from (2), we get : a = 23$
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:
$f (x) = x^{4} + a x^{3} - 7 x^{2} - 8 x + b$
Now,
$x + 2 = 0 \Rightarrow x = - 2$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x + 2) if f (- 2) = 0$ .
Thus, we have:
$f (- 2) = [{(- 2)}^{4} + a \times {(- 2)}^{3} - 7 \times {(- 2)}^{2} - 8 \times (- 2) + b] = (16 - 8 a - 28 + 16 + b) = (4 - 8 a + b)$
∴ $f (- 2) = 0 \Rightarrow 8 a - b = 4 . . . (1)$
Also,
$x + 3 = 0 \Rightarrow x = - 3$
By the factor theorem, we can say:
$f (x) will be exactly divisible by (x + 3) if f (- 3) = 0$ .
Thus, we have:
$f (- 3) = [{(- 3)}^{4} + a \times {(- 3)}^{3} - 7 \times {(- 3)}^{2} - 8 \times (- 3) + b] = (81 - 27 a - 63 + 24 + b) = (42 - 27 a + b)$
∴ $f (- 3) = 0 \Rightarrow 27 a - b = 42 . . . (2)$
$Subtracting (1) from (2), we get : \Rightarrow 19 a = 38 \Rightarrow a = 2$
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 $(x - \frac{1}{2})$ 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

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

Using factor theorem, we have

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

$4 p + r = p + 4 r \Rightarrow 4 p - p = 4 r - r \Rightarrow 3 p = 3 r \Rightarrow p = r$

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.

$x^{2} - 3 x + 2 = x^{2} - 2 x - x + 2 = x (x - 2) - 1 (x - 2) = (x - 1) (x - 2)$
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

$f (2) = 0 \Rightarrow 2 \times 2^{4} - 5 \times 2^{3} + 2 \times 2^{2} - 2 - 3 + k = 0 \Rightarrow 32 - 40 + 8 - 5 + k = 0 \Rightarrow - 5 + k = 0 \Rightarrow k = 5$
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.

RS AGGARWAL CLASS 9 Chapter 2 POLYNOMIALS EXERCISE 2D