Solve left(x-3right)^2+left(3x+1right)^2-10(x-1)(x+1)

$${ \left(x-3 \right) }^{ 2 } + { \left(3x+1 \right) }^{ 2 } -10(x-1)(x+1)$$

Evaluate

$$20$$

Solution Steps

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

$$x^{2}-6x+9+\left(3x+1\right)^{2}-10\left(x-1\right)\left(x+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$x^{2}-6x+9+9x^{2}+6x+1-10\left(x-1\right)\left(x+1\right)$$

Combine x^{2} and 9x^{2} to get 10x^{2}.

$$10x^{2}-6x+9+6x+1-10\left(x-1\right)\left(x+1\right)$$

Combine -6x and 6x to get 0.

$$10x^{2}+9+1-10\left(x-1\right)\left(x+1\right)$$

Add 9 and 1 to get 10.

$$10x^{2}+10-10\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply -10 by x-1.

$$10x^{2}+10+\left(-10x+10\right)\left(x+1\right)$$

Use the distributive property to multiply -10x+10 by x+1 and combine like terms.

$$10x^{2}+10-10x^{2}+10$$

Combine 10x^{2} and -10x^{2} to get 0.

$$10+10$$

Add 10 and 10 to get 20.

$$20$$

Factor

$$2^{2}\times 5$$