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  • The polynomial p(x) = x^{4} – 2x^{3} + 3x^{2} – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

The polynomial p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

Answers (1)

Value of a is 5

Remainder is 62

Solution

Given: p(x) = x4 – 2x3 + 3x2 – ax + 3a – 7

According to remainder theorem when p(x)  is divided by (x+a)  then the remainder is p(-a) .

When we divide p(x) by (x + 1) then according to remainder theorem remainder is p(–1)

\\p(-1)=(-1)^{4}-2(-1)^{3}+3(-1)^{2}-a(-1)+3a-7\\ =1+2+3+a+3a-7\\ =4a-1

According to question p(-1)=19
\Rightarrow 4a-1=19\\ \Rightarrow 4a=19+1\\ \Rightarrow 4a=20\\ \Rightarrow a=\frac{20}{4}=5\\ \therefore p(x)=x^{4}-2x^{3}+3x^{2}-5x+3(5)-7\\ =x^{4}-2x^{3}+3x^{2}-5x+8

When we divide p(x) by x + 2 then we get the remainder p(–2)

p(-2)=(-2)^{4}-2(-2)^{3}+3(-2)^{2}-5(-2)+8\\ =16+16+12+10+8\\ =62

Hence, value of a is 5.

Remainder is 62.

Posted by

infoexpert24

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