Question

Show that $(x+4),(x-3)$ and $(x-7)$ are factors of $x^3-6x^2-19x+84$

Answer 1

Factor theorem:

If $P\left(x\right)$ ia a polynomial, $x-a$ is a factor of $P\left(x\right)$ then $P\left(a\right)=0$ is remainder.

Let $f\left(x\right)=x^3-6x^2-19x+84$ be the given polynomial.

Lets check the, $(x+4),(x-3)$ and $(x-7)$ are factors or not.

(i) $(x+4)$

Put $x+4=0$

$\Rightarrow x=-4$

$f\left(x\right)=x^3-6x^2-19x+84$

$\Rightarrow f(-4)=(-4{)}^3-6(-4{)}^2-19(-4)+84$

$\Rightarrow f(-4)=-64-6\left(16\right)+76+84$

$\Rightarrow f(-4)=-64-96+76+84$

$\Rightarrow f(-4)=160-160$

$\Rightarrow f(-4)=0$

By the factor theorem, $(x+4)$ is a factor of $f\left(x\right)=x^3-6x^2-19x+84$.

(ii) $(x-3)$

Put $x-3=0$

$\Rightarrow x=3$

$f\left(x\right)=x^3-6x^2-19x+84$

$\Rightarrow f\left(3\right)=(3{)}^3-6(3{)}^2-19\left(3\right)+84$

$\Rightarrow f\left(3\right)=27-6\left(9\right)-57+84$

$\Rightarrow f\left(3\right)=27-54-57+84$

$\Rightarrow f\left(3\right)=111-111$

$\Rightarrow f\left(3\right)=0$

By the factor theorem, $(x-3)$ is a factor of $f\left(x\right)=x^3-6x^2-19x+84$.

(iii) $(x-7)$

Put $x-7=0$

$\Rightarrow x=7$

$f\left(x\right)=x^3-6x^2-19x+84$.

$\Rightarrow f\left(7\right)=(7{)}^3-6(7{)}^2-19\left(7\right)+84$

$\Rightarrow f\left(7\right)=343-6\left(49\right)-133+84$

$\Rightarrow f\left(7\right)=343-294-133+84$

$\Rightarrow f\left(7\right)=427-427$

$\Rightarrow f\left(7\right)=0$

By the factor theorem, $(x-7)$ is a factor of $f\left(x\right)=x^3-6x^2-19x+84$.