Answer to Question #161578 in Algebra for Samir khan

Question

Factor the following equation a) f(x) = (x ^ 2 - 36)/(x ^ 2 + 3x + 2) f(x) = 2x ^ 3 + 3x ^ 2 - 3x - 2

Answer 1

a) Consider the function $f(x)=\frac{x^2-36}{x^2+3x+2}$

Factor the numerator $x^2-36$ using $a^2-b^2=(a+b)(a-b)$ as,

$x^2-36=(x+6)(x-6)$

Factorize the denominator as,

$x^2+3x+2=x^2+2x+x+2=(x+2)(x+1)$

Therefore, the function in factored form is $f(x)=\frac{(x+6)(x-6)}{(x+2)(x+1)}$

b) Consider the function $f(x)=2x^3+3x^2-3x-2$

For $x=1,f(1)=2+3-3-2=0$ , so $x-1$ is a factor of the function.

Expand the function to find the factor $x-1$ as,

$f(x)=2x^3-2x^2+5x^2-5x+2x-2$

$=2x^2(x-1)+5x(x-1)+2(x-1)$

$=(x-1)(2x^2+5x+2)$

$=(x-1)(2x^2+4x+x+2)$

$=(x-1)(2x(x+2)+1(x+2))$

$=(x-1)(2x+1)(x+2)$

Therefore, the factored form of the function is $f(x)=(x-1)(2x+1)(x+2)$ .