RS Aggarwal Solutions: Exercise 2D - Polynomials | Extra Documents & Tests for Class 9 PDF Download

Q.1. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x³ – 8, g(x) = x – 2
Ans.
Let:
p(x) = x³ – 8
Now,
g(x) = 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(2)=(2³ − 8)=0
Hence, (x − 2) is a factor of the given polynomial.

Q.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
Ans.
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:
p(3) = (2 × 3³ − 7 × 3² − 24 × 3 − 45)
= (54 + 63 − 72 − 45)
= 0
Hence, (x − 3) is a factor of the given polynomial

Q.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
Ans.
Let:
p(x) = 2x⁴ + 9x³ + 6x² – 11x – 6
Here,
x − 1 = 0 ⇒ x=1
By the factor theorem, (x − 1) is a factor of the given polynomial if p(1) = 0.
Thus, we have:
p(1)=(2×1⁴ + 9 × 1³ + 6 × 1² − 11 × 1 − 6)
= (2 + 9 + 6 − 11 −6)
= 0
Hence, (x − 1) is a factor of the given polynomial.

Q.4. Using factor theorem, show that g(x) is a factor of p(x), when
p(x) = x⁴ – x² – 12, g(x) = x + 2
Ans.
Let:
p(x) = x⁴ – x² – 12
Here,
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(−2) = [(−2)⁴ − (−2)² − 12]
= (16 − 4 − 12)
=0
Hence, (x + 2) is a factor of the given polynomial.

Q.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
Ans.
p(x)=69+11x−x²+x³
g(x) = x + 3
Putting x = −3 in p(x), we get
p(−3) = 69 + 11 × (−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).

Q.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
Ans.
Let:
p(x) = 2x³ + 9x² − 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:
p(−5) = [2 × (−5)³ +9 × (−5)² −11 × (−5)−30]
= (−250 + 225 + 55 − 30)
= 0
Hence, (x + 5) is a factor of the given polynomial.

Q.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

Ans.
Let:
p(x)=2x⁴ + x³ − 8x² − x + 6
Here,
2x − 3 = 0 ⇒ x =
By the factor theorem, (2x − 3) is a factor of the given polynomial if p= 0
Thus, we have:


= 0
Hence, (2x − 3) is a factor of the given polynomial.

Q.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
Ans.
p(x) = 3x³ + x² − 20x + 12
g(x) = 3x − 2 = 3
Putting in p(x), we get


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

Q.9. Using factor theorem, show that g(x) is a factor of p(x), when

p(x) = 7x² −4√2x − 6, g(x) = x − √2

Ans.
Let:
p(x)=7x²−4√2x − 6
Here,
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(√2) = [7×(√2)² −4√2 × √2 −6]
=(14 − 8 − 6)
= 0
Hence, (x − √2) is a factor of the given polynomial.

Q.10. Using factor theorem, show that g(x) is a factor of p(x), when
p(x)=2√2x² + 5x + √2, g(x) =x + √2
Ans.
Let:
p(x)=2√2x² + 5x + √2

Here,
x + √2 = 0 ⇒ x = −√2
By the factor theorem, (x+√2) will be a factor of the given polynomial if p(−√2) = 0.
Thus, we have:
p(−√2) = [2√2 × (−√2)² + 5 × (−√2)+√2]
=(4√2−5√2+√2)

= 0
Hence, (x + √2) is a factor of the given polynomial.

Q.11. Show that (p – 1) is a factor of (p¹⁰ – 1) and also of (p¹¹ – 1).
Ans.
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).

Q.12. Find the value of k for which (x − 1) is a factor of (2x³ + 9x² + x + k).
Ans.

Let:

f(x)=2x³ + 9x² + x + k

(x−1) is a factor of f(x) = 2x³ + 9x² + x + k.
⇒ f(1) = 0
⇒2 × 1³ + 9 × 1² + 1 + k = 0
⇒12 + k = 0
⇒k= −12
Hence, the required value of k is −12.

Q.13. Find the value of a for which (x − 4) is a factor of (2x³ − 3x² − 18x + a).
Ans.
Let:

f(x)=2x³ − 3x² − 18x + a
(x−4) is a factor of f(x)=2x³ − 3x² − 18x + a.
⇒ f(4) = 0
⇒2 × 4³ − 3 × 4² − 18 × 4 + a = 0
⇒8 + a = 0 ⇒ a = −8
Hence, the required value of a is −8.

Q.14. Find the value of a for which (x + 1) is a factor of (ax³ + x² – 2x + 4a – 9).
Ans.
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
⇒a×(−1)³+(−1)²−2×(−1) + 4a − 9 = 0
⇒− a + 1 + 2+ 4a −9 = 0
⇒ 3a −6 = 0
⇒ 3a = 6
⇒a = 2
Thus, the value of a is 2.

Q.15. Find the value of a for which (x + 2a) is a factor of (x⁵ – 4a² x³ + 2x + 2a +3).
Ans.

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
⇒(−2a)⁵−4a²×(−2a)³+2×(−2a)+2a+3=0
⇒−32a⁵−4a²×(−8a³)+2×(−2a)+2a+3=0
⇒−32a⁵+32a⁵−4a+2a+3=0
⇒ −2a + 3 = 0
⇒ 2a = 3
⇒
Thus, the value of a is 3/2.

Q.16. Find the value of m for which (2x – 1) is a factor of (8x⁴ + 4x³ – 16x² + 10x + m).
Ans.
Let f(x) = 8x⁴ + 4x³ – 16x² + 10x + m
It is given that (2x−1) = 2 is a factor of f(x).
Using factor theorem,  we have



⇒2 + m = 0
⇒m = −2

Thus, the value of m is −2.

Q.17. Find the value of a for which the polynomial (x⁴ − x³ − 11x² − x + a) is divisible by (x + 3).
Ans.
Let:
f(x) = x⁴ − x³ − 11x² − x + a
Now,
x + 3 = 0 ⇒ 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×(−3)²−(−3)+a]
=(81 + 27 − 99 + 3 + a)
=12 + a
Also,
f( − 3) = 0
⇒ 12 + a = 0
⇒ a = −12

Hence, f(x) is exactly divisible by (x+3) when a is −12.

Q.18. Without actual division, show that (x³ − 3x² − 13x + 15) is exactly divisible by (x² + 2x − 3).
Let:
f(x)=x³ − 3x² −13 x + 15
And,
g(x) = x² + 2x − 3
=x² + x − 3x −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) = (1³ − 3 × 1² − 13 × 1 + 15)
=(1 − 3 − 13 + 15)
= 0
And,
f(−3)=[(−3)³−3×(−3)²−13×(−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 by x² + 2x − 3.

Q.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.
Ans.
Let:
f(x)=x³ + ax² + bx + 6
(x−2) is a factor of f(x)=x³ + ax² + bx + 6.
⇒f(2) = 0
⇒2³ + a × 2² + b × 2 + 6 = 0
⇒14 + 4a + 2b =0
⇒4a + 2b = −14
⇒ 2a + b= −7         ...(1)
Now,
x − 3 = 0 ⇒ x = 3
By the factor theorem, we can say:
When f(x) will be divided by (x−3), 3 will be its remainder.
⇒ f(3) = 3
Now,
f(3)=3³ + a × 3² + b × 3 + 6
=(27 + 9a + 3b + 6)
= 33 + 9a + 3b
Thus, we have:
f (3) = 3
⇒33 + 9a + 3b = 3
⇒9a + 3b = −30
⇒3a + 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

Q.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).
Ans.
Let:
f(x) =x³ − 10 x² + ax + b
Now,
x − 1 = 0 ⇒ 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)=1³ − 10 × 1² + a × 1 + b
=(1 − 10 + a + b)
= −9 + a + b
∴ f(1) = 0 ⇒ a + b = 9 ...(1)
Also,
x − 2 = 0 ⇒ 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)=2³ − 10 × 2² + a × 2 + b
=(8 − 40 + 2a + b)
= − 32 + 2a + b
∴ f(2) = 0 ⇒ 2a+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

Q.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).
Ans.
Let:
f(x)=x⁴ + ax³ − 7x² − 8x + b
Now,
x + 2 = 0 ⇒ 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×(−2)³−7×(−2)²−8×(−2)+b]
=(16 − 8a − 28 +16+ b)
=(4 − 8a + b)

∴ f(−2) = 0 ⇒ 8a − b = 4      ...(1)
Also,
x + 3 = 0 ⇒ 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×(−3)³−7×(−3)²−8×(−3)+b
=(81 −27a − 63 + 24 + b)
= (42 − 27 a + b)
∴ f(−3) = 0 ⇒ 27a − b = 42   ...(2)
Subtracting (1) from (2), we get:
⇒19a = 38
⇒ a = 2
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

Q.22. If both (x – 2) and are factors of px² + 5x + r, prove that p = r.
Ans.
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
⇒p × 2² + 5 × 2 + r = 0
⇒ 4p + r= −10 .....(1)
Also,is a factor of f(x).
Using factor theorem,  we have


From (1) and (2), we have
4p + r = p + 4r
⇒ 4p − p= 4r − r
⇒3p = 3r
⇒ p = r

Q.23. Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.
Ans.
Let f(x) = 2x⁴ – 5x³ + 2x² – x + 2 and g(x) = x² – 3x + 2.
x2 − 3x + 2
=x² − 2x − 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.

Q.24. What must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2)?
Ans.

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
⇒2×2⁴−5×2³+2×2²−2−3+k=0
⇒32 − 40 + 8 − 5 + k = 0

⇒ −5 + k = 0
⇒k = 5
Thus, 5 must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2).

Q.25. What must be subtracted from (x⁴ + 2x³ – 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x – 3)?
Ans.
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).

Q.26. Use factor theorem to prove that (x + a) is a factor of (xⁿ + aⁿ) for any odd positive integer.
Ans.
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.