Question

Use the remainder theorem to find P (1) for P(x) = -2x⁴+3x³ + 6x² -7. Specifically, give the quotient and the remainder for the associated division and the value of P (1).

Answer 1

Step 1

Given;

$P(x)=-2x^4+3x^3+6x^2-7$

Required; To find P(1)

Step 2

Set x=1

$\begin{gathered} P(1)=-2(1)^4+3(1)^3+6(1)^2-7 \\ =-2\cdot \:1+3\left(1\right)^3+6\left(1\right)^2-7 \\ =-2\cdot \:1+3\cdot \:1+6\cdot \:1-7 \\ =-2+3\cdot \:1+6\cdot \:1-7 \\ =-2+3+6\cdot \:1-7 \\ =-2+3+6-7 \\ =0 \end{gathered}$

Using the factor remainder theorem;

Hence;

$\begin{gathered} Quotient=-2x^3+x^2+7x+7 \\ Remainder=0 \\ P(1)=0 \end{gathered}$