Solve (x^2-3x)^2-14(x^2-3x)+40=

$$( x ^ { 2 } - 3 x ) ^ { 2 } - 14 ( x ^ { 2 } - 3 x ) + 40 =$$

Evaluate

$$\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-3x\right)^{2}.

$$\left(x^{2}\right)^{2}-6x^{2}x+9x^{2}-14\left(x^{2}-3x\right)+40$$

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

$$x^{4}-6x^{2}x+9x^{2}-14\left(x^{2}-3x\right)+40$$

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

$$x^{4}-6x^{3}+9x^{2}-14\left(x^{2}-3x\right)+40$$

Use the distributive property to multiply -14 by x^{2}-3x.

$$x^{4}-6x^{3}+9x^{2}-14x^{2}+42x+40$$

Combine 9x^{2} and -14x^{2} to get -5x^{2}.

$$x^{4}-6x^{3}-5x^{2}+42x+40$$

Expand

$$x^{4}-6x^{3}-5x^{2}+42x+40$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-3x\right)^{2}.

$$\left(x^{2}\right)^{2}-6x^{2}x+9x^{2}-14\left(x^{2}-3x\right)+40$$

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

$$x^{4}-6x^{2}x+9x^{2}-14\left(x^{2}-3x\right)+40$$

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

$$x^{4}-6x^{3}+9x^{2}-14\left(x^{2}-3x\right)+40$$

Use the distributive property to multiply -14 by x^{2}-3x.

$$x^{4}-6x^{3}+9x^{2}-14x^{2}+42x+40$$

Combine 9x^{2} and -14x^{2} to get -5x^{2}.

$$x^{4}-6x^{3}-5x^{2}+42x+40$$