Question

Q2. p(x) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by (x - 2) and q(x) = 3x³ - 2x² + x - 1 is divided by (x - 1). So, twice the sum of the remainders is:

(A) 21
(B) 35
(C) 54
(D) 40​

Answer 1

Hint

• Remainder theorem

If we divide a polynomial by a linear polynomial, the remainder can be found by substituting the zero of the divisor.

Solution

$p(2)$ is the remainder when $p(x)$ is divided by $(x-2)$. Similarly, $q(1)$ is the remainder when $q(x)$ is divided by $(x-1)$.

So, the remainders are $p(2)=19$ and $q(1)=1$ respectively.

Twice the sum of the remainders is $\boxed{40}$. It is option (D).

Answer 2

Given :-

p(x) = 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by (x - 2) and q(x) = 3x³ - 2x² + x - 1 is divided by (x - 1)

To Find :-

Twice the sum of the remainders is:

Solution :-

x - 2 = 0

x = 2

p(2) = 2(2)⁴ - 3(2)³ + 2(2)² - 2(2) - 1

p(-2) = 2(16) - 3(8) + 2(4) - (4) - 1

p(-2) = 32 - 24 + 8 + 4 - 1

p(-2) = 32 - 24 + 8 + 4 - 1

p(-2) = 8 + 8 + 4 - 1

p(-2) = 20 - 1

p(-2) = 19

Now

x - 1 = 0

x = 0 + 1

x = 1

q(1) = 3(1)³ - 2(1)² + 1 - 1

q(1) = 3 × 1 - 2 × 1 + 1 - 1

q(1) = 3 - 2 + 1 - 1

q(1) = 4 - 3

q(1) = 1

Now

Twice the sum

2(19 + 1)

2(20)

40

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