Question

Find The Integer Without Actual Division Prove That $x^3-3x^2-13x+15$ Is Exactly Divisible By $x^2+2x-3$.

Answer 1

Step 1: Find the factors

Given Polynomials: $x^3-3x^2-13x+15$ and$x^2+2x-3$.

Let $f\left(x\right)=x^3-3x^2-13x+15$ and $g\left(x\right)=x^2+2x-3$

Then,

$$\begin{gathered} g\left(x\right)=x^2+2x-3 \\ =x^2+3x-x-3 \\ =x(x+3)-1(x+3) \\ =(x-1)(x+3) \end{gathered}$$

This means $(x-1)$ and $(x+3)$ are factors of$g\left(x\right)$.

Step 2: Find the values of$f\left(x\right)$.

At, $x=1$ the value of $f\left(x\right)$ is,

$$\begin{gathered} f\left(1\right)={\left(1\right)}^3-3{\left(1\right)}^2-13\left(1\right)+15 \\ =1-3-13+15 \\ =-15+15 \\ =0 \end{gathered}$$

So, by factor theorem we can say $(x-1)$ is a factor of$f\left(x\right)$.

At $x=-3$ the value of $f\left(x\right)$ is,

$$\begin{gathered} f(-3)={(-3)}^3-3{(-3)}^2-13(-3)+15 \\ =-27-3\left(9\right)+39+15 \\ =-27-27+54 \\ =-54+54 \\ =0 \end{gathered}$$

So, by factor theorem we can say $(x-\left(-3\right))=(x+3)$ is a factor of$f\left(x\right)$.

As both $(x-1)$ and $(x+3)$ are factors of the given polynomials $f\left(x\right)$ and $g\left(x\right)=(x-1)(x+3)$

So we can say $x^3-3x^2-13x+15$ Is Exactly Divisible By$x^2+2x-3$.

Hence proved.