Solve y=(x^2-6x)^2+12(x^2-6x)+30quad0

$$y = ( x ^ { 2 } - 6 x ) ^ { 2 } + 12 ( x ^ { 2 } - 6 x ) + 30 \quad 0$$

Solve for y

$$y=x\left(x-6\right)\left(x^{2}-6x+12\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-6x\right)^{2}.

$$y=\left(x^{2}\right)^{2}-12x^{2}x+36x^{2}+12\left(x^{2}-6x\right)+30\times 0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$y=x^{4}-12x^{2}x+36x^{2}+12\left(x^{2}-6x\right)+30\times 0$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$y=x^{4}-12x^{3}+36x^{2}+12\left(x^{2}-6x\right)+30\times 0$$

Use the distributive property to multiply 12 by x^{2}-6x.

$$y=x^{4}-12x^{3}+36x^{2}+12x^{2}-72x+30\times 0$$

Combine 36x^{2} and 12x^{2} to get 48x^{2}.

$$y=x^{4}-12x^{3}+48x^{2}-72x+30\times 0$$

Multiply 30 and 0 to get 0.

$$y=x^{4}-12x^{3}+48x^{2}-72x+0$$

Anything plus zero gives itself.

$$y=x^{4}-12x^{3}+48x^{2}-72x$$