Solve 3(x-4)^2-(3y-4)^2

$$3 ( x - 4 ) ^ { 2 } - ( 3 y - 4 ) ^ { 2 }$$

Expand

$$3x^{2}-24x-9y^{2}+24y+32$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$3\left(x^{2}-8x+16\right)-\left(3y-4\right)^{2}$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$3x^{2}-24x+48-\left(3y-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-4\right)^{2}.

$$3x^{2}-24x+48-\left(9y^{2}-24y+16\right)$$

To find the opposite of 9y^{2}-24y+16, find the opposite of each term.

$$3x^{2}-24x+48-9y^{2}+24y-16$$

Subtract 16 from 48 to get 32.

$$3x^{2}-24x+32-9y^{2}+24y$$

Evaluate

$$3\left(x-4\right)^{2}-\left(3y-4\right)^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$3\left(x^{2}-8x+16\right)-\left(3y-4\right)^{2}$$

Use the distributive property to multiply 3 by x^{2}-8x+16.

$$3x^{2}-24x+48-\left(3y-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-4\right)^{2}.

$$3x^{2}-24x+48-\left(9y^{2}-24y+16\right)$$

To find the opposite of 9y^{2}-24y+16, find the opposite of each term.

$$3x^{2}-24x+48-9y^{2}+24y-16$$

Subtract 16 from 48 to get 32.

$$3x^{2}-24x+32-9y^{2}+24y$$