Question

Question 2
The polynomial $p\left(x\right)=x^4–2x^3+3x^2–ax+3a–7$ when divided by x + 1 leaves the remainder 19. Find the value of a. Also , find the remainder when p(x) is divided by x + 2.

Answer 1

Given,
$p\left(x\right)=x^4–2x^3+3x^2–ax+3a–7$
When we divide $p\left(x\right)byx+1$, then we get the remainder as 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 the question, p(-1) = 19
$\Rightarrow 4a–1=19$
$\Rightarrow 4a=20$
$\therefore a=5$
$\therefore$ Required polynomial $=x^4–2x^3+3x^2–5x+3\left(5\right)–7$
$=x^4–2x^3+3x^2–5x+15–7$
$=x^4–2x^3+3x^2–5x+8$
When we divide p(x) by x + 2, we get the remainder as p(-2).
Now, $p(-2)=(-2{)}^4–2(-2{)}^3+3(-2{)}^2–5(-2)+8$
$=16+16+12+10+8=62$