Solve (r-5)(r+5)+(r-5)(r+2)

$$( r - 5 ) ( r + 5 ) + ( r - 5 ) ( r + 2 )$$

Evaluate

$$\left(r-5\right)\left(2r+7\right)$$

Solution Steps

Consider \left(r-5\right)\left(r+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$r^{2}-5^{2}+\left(r-5\right)\left(r+2\right)$$

Calculate 5 to the power of 2 and get 25.

$$r^{2}-25+\left(r-5\right)\left(r+2\right)$$

Apply the distributive property by multiplying each term of r-5 by each term of r+2.

$$r^{2}-25+r^{2}+2r-5r-10$$

Combine 2r and -5r to get -3r.

$$r^{2}-25+r^{2}-3r-10$$

Combine r^{2} and r^{2} to get 2r^{2}.

$$2r^{2}-25-3r-10$$

Subtract 10 from -25 to get -35.

$$2r^{2}-35-3r$$

Expand

$$2r^{2}-3r-35$$

Solution Steps

Consider \left(r-5\right)\left(r+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$r^{2}-5^{2}+\left(r-5\right)\left(r+2\right)$$

Calculate 5 to the power of 2 and get 25.

$$r^{2}-25+\left(r-5\right)\left(r+2\right)$$

Apply the distributive property by multiplying each term of r-5 by each term of r+2.

$$r^{2}-25+r^{2}+2r-5r-10$$

Combine 2r and -5r to get -3r.

$$r^{2}-25+r^{2}-3r-10$$

Combine r^{2} and r^{2} to get 2r^{2}.

$$2r^{2}-25-3r-10$$

Subtract 10 from -25 to get -35.

$$2r^{2}-35-3r$$