Question

Identify all of the roots of g(x)=(x^2+3x-4)(x^2-4x+29)

Answer 1

Answer:

The roots of $\bold{\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)}$ is -4 and 1

Solution:

From question, given that $\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)$

To factorize the first expression that is $\left(x^{2}+3 x-4\right)$ ,follow the below steps:

$\left(x^{2}+3 x-4\right)$ ---- eqn (1)

“3x” can be rewritten as 4x-x. Now the above equation becomes,

$=\left(x^{2}+4 x-x-4\right)$

By taking x as common from $x^{2}+4 x \text { and }$ -1 from -x-4 the above equation becomes,

=x (x+4) - 1(x+4)

=(x+4)(x-1)

Hence the factors of $\left(x^{2}+3 x-4\right) \text { is }(x+4)(x-1)$    

so the roots of $\left(x^{2}+3 x-4\right)$ is -4 and 1.

Now to factorize the second expression that is $\left(x^{2}-4 x+29\right)$ ,follow the below steps:

$\left(x^{2}+3 x-4\right)$  --- eqn (2)

Equation (2) cannot be factorized (imaginary roots)

Hence the roots of $\left(x^{2}+3 x-4\right)\left(x^{2}-4 x+29\right)$ is -4 and 1.