Solve left(5x-2right)^2-left(3x+1right)^2-x^2-60

$${ \left(5x-2 \right) }^{ 2 } - { \left(3x+1 \right) }^{ 2 } - { x }^{ 2 } -60$$

Evaluate

$$\left(x-3\right)\left(15x+19\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-\left(3x+1\right)^{2}-x^{2}-60$$

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

$$25x^{2}-20x+4-\left(9x^{2}+6x+1\right)-x^{2}-60$$

To find the opposite of 9x^{2}+6x+1, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}-6x-1-x^{2}-60$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-20x+4-6x-1-x^{2}-60$$

Combine -20x and -6x to get -26x.

$$16x^{2}-26x+4-1-x^{2}-60$$

Subtract 1 from 4 to get 3.

$$16x^{2}-26x+3-x^{2}-60$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-26x+3-60$$

Subtract 60 from 3 to get -57.

$$15x^{2}-26x-57$$

Expand

$$15x^{2}-26x-57$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-\left(3x+1\right)^{2}-x^{2}-60$$

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

$$25x^{2}-20x+4-\left(9x^{2}+6x+1\right)-x^{2}-60$$

To find the opposite of 9x^{2}+6x+1, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}-6x-1-x^{2}-60$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-20x+4-6x-1-x^{2}-60$$

Combine -20x and -6x to get -26x.

$$16x^{2}-26x+4-1-x^{2}-60$$

Subtract 1 from 4 to get 3.

$$16x^{2}-26x+3-x^{2}-60$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}-26x+3-60$$

Subtract 60 from 3 to get -57.

$$15x^{2}-26x-57$$