**Previous Exercise – Polynomials – Exercise 8.3 – Class X**

#### Polynomials – Exercise 8.4

- In each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.

(i) p(x) = x^{3} – 3x^{2} + 6x – 20 ; g(x) = x – 2

(ii) p(x) = 2x^{4} + x^{3} + 4x^{2} – x – 7 ; g(x) = x + 2

(iii) p(x) = 3x^{4} + 3x^{2} – 2x^{2} – 9x – 12 ; g(x) = x – ^{1}/_{2}

(iv) p(x) = 3x^{3} + x^{2} – 20x + 12 ; g(x) = 3x – 2

(v) p(x) = 2x^{4} + 3x^{3} – 2x^{2} – 9x – 12 ; g(x) = x^{2} – 3

- Find the valule of a if (x – 5) is a factor of (x
^{3}– 3x^{2}+ ax – 10) - If (x
^{3}+ ax – bx + 10) is divisible by x^{2}– 3x + 2 , find the values of a and b - If both (x – 2) and (x –
^{1}/_{2})are factors of (ax^{2}+ 5x + b), show that a = b

#### Polynomials – Exercise 8.4 –Solutions:

- I
**n each of the following cases, use factor theorem to find whether g(x) is a factor of the polynomials p(x) or not.**

**(i) p(x) = x ^{3} – 3x^{2} + 6x – 20 ; g(x) = x – 2**

Solution:

Let p(x) = x^{3} – 3x^{2} + 6x – 20

By factor theorem, (x – 2) is a factor of p(x) if p(2) = 0

p(2) = 2^{3} – 3(2)^{2} + 6(2) – 20

= 8 – 12 + 12 – 20

= – 12

⸫ (x – 2) is not a factor of p(x) = x^{3} – 3x^{2} + 6x – 20

**(ii) p(x) = 2x ^{4} + x^{3} + 4x^{2} – x – 7 ; g(x) = x + 2**

Solution:

Let p(x) = 2x^{4} + x^{3} + 4x^{2} – x – 7

By factor theorem, (x + 2) is a factor of p(x) if p(-2) = 0

p(-2) = 2(-2)^{4} + (-2)^{3} + 4(-2)^{2} – (-2) – 7

= 32 – 8 + 16 + 2 – 7

= 35

⸫ (x + 2) is not a factor of p(x) = 2x^{4} + x^{3} + 4x^{2} – x – 7

** (iii) p(x) = 3x ^{4} + 3x^{3} – 2x^{2} – 9x – 12 ; g(x) = x – ^{1}/_{2}**

Solution:

Let p(x) = 3x^{4} + 3x^{3} – 2x^{2} – 9x – 12

By factor theorem, (x – ^{1}/_{2}) is a factor of p(x) if p(^{1}/_{2}) = 0

p(^{1}/_{2}) = 3(^{1}/_{2})^{4} + 3(^{1}/_{2})^{3} – 2(^{1}/_{2})^{2} – 9(^{1}/_{2}) – 12

= 3(^{1}/_{16}) + 3(^{1}/_{8}) – 2(^{1}/_{4}) – 9(^{1}/_{2}) – 12

= – ^{263}/_{16}

⸫ (x – ^{1}/_{2}) is not a factor of p(x) = 3x^{4} + 3x^{2} – 2x^{2} – 9x – 12

** (iv) p(x) = 3x ^{3} + x^{2} – 20x + 12 ; g(x) = 3x – 2**

Solution:

Let p(x) = 3x^{3} + x^{2} – 20x + 12

By factor theorem, (3x – 2) is a factor of p(x) if p(^{2}/_{3}) = 0

p(^{2}/_{3}) = 3(^{2}/_{3})^{3} + (^{2}/_{3})^{2} – 20(^{2}/_{3}) + 12

= 0

⸫ (3x – 2) is a factor of p(x) = 3x^{3} + x^{2} – 20x + 12

** (v) p(x) = 2x ^{4} + 3x^{3} – 2x^{2} – 9x – 12 ; g(x) = x^{2} – 3**

Solution:

Let p(x) = 2x^{4} + 3x^{3} – 2x^{2} – 9x – 12

By factor theorem, (x^{2} – 3) is a factor of p(x) if p(√3) = 0

p (√3) = 2 (√3)^{4} + 3 (√3)^{3} – 2 (√3)^{2} – 9 (√3) – 12

= 2x3x3 + 3×3√3 – 2×3 – 9√3 – 12

= 18 + 9√3 – 6 – 9√3 – 12

= 0

⸫ (x^{2} – 3) is a factor of p(x) = 2x^{4} + 3x^{3} – 2x^{2} – 9x – 12

**Find the value of a if (x – 5) is a factor of (x**^{3}– 3x^{2}+ ax – 10)

Solution:

Let p(x) = x^{3} – 3x^{2} + ax – 10

By factor theorem, (x – 5) is a factor of p(x) if p(5) = 0

p(x) = x^{3} – 3x^{2} + ax – 10 = 0

5^{3} – 3(5^{2}) + a(5) – 10 = 0

125 – 75 + 5a – 10 = 0

40 + 5a = 0

5a = – 40

a = –^{40}/_{5} = – 8

**If (x**^{3}+ ax^{2}– bx + 10) is divisible by x^{2}– 3x + 2 , find the values of a and b

Solution:

p(x) = x^{3} + ax^{2} – bx + 10

g(x) = x^{2} – 3x + 2

g(x) = x^{2} –x – 2x + 2

g(x) = x(x – 1)-2(x – 1)

g(x) = (x – 1)(x – 2)

p(x) = (x^{3} + ax^{2} – bx + 10) is divisible by g(x) = (x – 1)(x – 2) , then we have p(2) = 0 and p(1) = 0

⇒ p(2) = p(1)

We have to find the values of a and b.

p(x) = x^{3} + ax^{2} – bx + 10

p(2) = (2)^{3} + a(2)^{2} – b(2) + 10 = 8 + 4a – 2b + 10 = 4a – 2b + 18 = 2a – b + 9 = 0

p(1) = (1)^{3} + a(1)^{2} – b(1) + 10 = 1 + a – b + 10 = a – b + 11 = 0

Since p(2) = p(1)

2a – b + 9 = a – b + 11

2a – a = 11 – 9

a = 2

Substitute the value of a in p(2) = 0

p(2) = a – b + 11 = 0

⇒ 2 – b + 11 = 0

-b = -11 – 2 = – 13

b = 13

Therefore, a = 2 and b = 13

**If both (x – 2) and (x –**^{1}/_{2})are factors of (ax^{2}+ 5x + b), show that a = b

Solution:

p(x) = ax^{2} + 5x + b

By factor theorem, (x – 2) and (x – ^{1}/_{2}) is a factor of p(x) then p(2) = p(^{1}/_{2}) = 0

p(2) = a(2)^{2} + 5(2) + b = 4a + b + 10 = 0

p(^{1}/_{2}) = a(^{1}/_{2})^{2} + 5(^{1}/_{2}) + b = ^{a}/_{4} + ^{5}/_{2} + b = 0

Since p(2) = p(^{1}/_{2})

4a + b + 10 = ^{a}/_{4} + ^{5}/_{2} + b

4a + 10 = ^{a}/_{4} + ^{5}/_{2}

4a – ^{a}/_{4} = ^{5}/_{2} – 10

^{16a – a}/_{4} = ^{5-20}/_{2}

^{15a}/_{4} = ^{15}/_{2}

^{a}/_{2} = 1

a = 2

Substitute the value of p(2) = 4a + b + 10 = 0

4a + b + 10 = 0

8 + b + 10 = 0

b = -18

**Next exercise – Polynomials – Exercise 8.5 – Class X**