Question

1) Find the remainder when $x^{3}$- $ax^{2}$+ 6x- a is divided by x-a. 2) What is the remainder when $x^{2018}$ + 2018 is divided by x – 1? 3) For what value of k is the polynomial p(x)=$2x^{3}-kx^{2}$+ 3x + 10 is exactly divisible by (x– 2)? 4) If two polynomials $2x^{3}+ax^{2}$+ 4x – 12 and $x^{3}+x^{2}$– 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.

Answer 1

1.

p(x) = x³ - ax² + 6x - a

g(x) = x - a

We need to find the remainder when p(x) is divided by g(x)

∴ p(g(x)) = (a)³ - a(a)² + 6(a) - a

⇒ p(g(x)) = a³ - a³ + 5a

⇒ p(g(x)) = 5a

Hence, The remainder is 5a

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2.

p(x) = x²⁰¹⁸ + 2018

g(x) = x - 1

Here, We need to find the remainder when p(x) is divided by g(x), which is equal to:

⇒ p(g(x)) = (1)²⁰¹⁸ + 2018

⇒ p(g(x)) = 1 + 2018

⇒ p(g(x)) = 2019

Hence, The remainder we would get is 2019.

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3.

p(x) = 2x³ - kx² + 3x + 10

g(x) = x - 2

We need to find a value of k such that p(x) would be exactly divisible by g(x), In other words, remainder would be 0.

⇒ p(g(x)) = 0

⇒ 2(2)³ - k(2)² + 3(2) + 10 = 0

⇒ 16 - 4k + 6 + 10 = 0

⇒ -4k = -32

⇒ k = 4

Hence, For k = 4, p(x) would be exactly divisible by g(x)

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4.

p(x) = 2x³ + ax² + 4x - 12

q(x) = x³ + x² - 2x + a

g(x) = x - 3

Given that, when p(x) is divided by g(x) and q(x) by g(x) , they leave the same remainder. so we need to find the value of a and the remainder as well.

⇒ p(g(x)) = q(g(x))

⇒ 2(3)³ + a(3)² + 4(3) - 12 = (3)³ + (3)² - 2(3) + a

⇒ 54 + 9a + 12 - 12 = 27 + 9 - 6 + a

⇒ 9a - a = 30 - 54

⇒ 8a = -24

⇒ a = -3

Now,

Remainder = p(g(x)) or q(g(x))

⇒ p(g(x))

⇒ 2(3)³ - 3 × 3² + 4(3) - 12

⇒ 54 - 27

⇒ 27

Remainder = 27

Verification:

⇒ q(g(x)) = 27

⇒ (3)³ + (3)² -2(3) - 3 = 27

⇒ 27 + 9 - 6 - 3 = 27

⇒ 30 - 3 = 27

⇒ 27 = 27

Hence, Proved.

Therefore, Value of a is -3 and the Remainder is 27.

Answer 2

Question:-

• Find the remainder when x³ - 3x² + 4x - 12 is divided by x - 3.

Answer: -

• Dividing x³ - 3x² + 4x - 12 by x - 3,

• Dividend: x³ - 3x² + 4x - 12

• Divisor : x - 3

Now, Put divisor as 0,

• x - 3 = 0

• .°. x = 3

Now,

• Let p (x) = x³ - 3x² + 4x - 12

Putting value of x,

• p(3) = (3)³ - 3(3)² + 4(3) - 12

• = 27 - 27 + 12 - 12

• = 0 + 0

• = 0

Thus, remainder = p(3) = 0

So, the remainder when x³ - 3x² + 4x - 12 is divided by x - 3 is 0.