Find the Roots (Zeros) y=(x-4)(x^2-12x+36)

$y = (x - 4) (x^{2} - 12 x + 36)$

Step 1

Set $(x - 4) (x^{2} - 12 x + 36)$ equal to $0$ .

$(x - 4) (x^{2} - 12 x + 36) = 0$

Step 2

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x^{2} - 12 x + 36 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x^{2} - 12 x + 36$ equal to $0$ and solve for $x$ .

Set $x^{2} - 12 x + 36$ equal to $0$ .

$x^{2} - 12 x + 36 = 0$

Solve $x^{2} - 12 x + 36 = 0$ for $x$ .

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$x^{2} - 12 x + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 x = 2 \cdot x \cdot 6$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 6$ .

${(x - 6)}^{2} = 0$

Set the $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $(x - 4) (x^{2} - 12 x + 36) = 0$ true.

$x = 4, 6$

Step 3