Let p(x) be a quadratic polynomial such that p(0)=1.

Question

Let $p(x)$ be a quadratic polynomial such that $p(0)=1$. If $p(x)$ leaves remainder 4 when divided by $x-1$, and remainder 6 when divided by $x+1$, then

Accepted Answer

Let, P(x) = ax2 + bx + c

As, P(0) = 1,

$$\therefore\,\,\,$$ a(0)2 + b(0) + c = 1

$$ \Rightarrow $$ $\,\,\,$c = 1$\therefore\,\,\,$$ P(x) = ax2 + bx + 1

If   P(x) is divided by x $$-$1,  remainder = 4$ \Rightarrow $$ $\,\,\,$$ P$$\left( 1 \right) = 4$$

$$\therefore\,\,\,$$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$$ \Rightarrow $$ $\,\,\,$P($-$1) = 6$\therefore\,\,\,$a$-$$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $$-$1$\therefore\,\,\,$P(x) = 4x2$-$x + 1

P(2) = 4(2)2$-$2 + 1 = 15

P($-$2) = 4 ($-$2)2$-$($-$

Solutions · 1