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Q.

What must be subtracted or added to p(x) = 8 $x^{4}$ + 14 $x^{3}$ – 2 $x^{2}$ + 8x – 12 so that 4 $x^{2}$ + 3x − 2 is a factor of p(x)?

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a

15x − 14

b

12x − 14

c

15x − 16

d

12x − 16

answer is A.

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Detailed Solution

Given p(x) = 8

x^{4}

+ 14

x^{3}

– 2

x^{2}

+ 8x – 12 and let q(x) = 4

x^{2}

+ 3x – 2.
We need to find what must be added or subtracted from p(x) so that q(x) becomes a factor of p(x). For this, we will find the remainder when p(x) is divided by q(x). As the remainder will be the added term in p(x), we will subtract it from p(x) to obtain the required answer.
Let us use long division method for dividing 8

x^{4}

+ 14

x^{3}

– 2

x^{2}

+ 8x – 12 by 4

x^{2}

+ 3x – 2.

2 x^{2} + 2 x - 1

4

x^{2}

+ 3x – 2 )

\underset{̲}{8 x^{4} + 14 x^{3} - 2 x^{2} + 8 x - 12}

- 8

x^{4}

+ 6

x^{3}

- 4

x^{2}

\underset{̲}{0 + 8 x^{3} + 2 x^{2} + 8 x - 12}

- (

8 x^{3} + 6 x^{2} - 4 x

)

\underset{̲}{0 - 4 x^{2} + 12 x - 12}

− (− 4

x^{2}

− 3x +2 )

\underset{̲}{0 + 15 x - 14}

So 15x−14 is the required remainder.
Hence, we will have to subtract 15x−14 from 8

x^{4}

+ 14

x^{3}

– 2

x^{2}

+ 8x – 12 so that 4

x^{2}

+ 3x – 2 become its factor.
So, option (1) is correct.

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