Solve (3x-4)(3x+4)+5(2x+1)(5-4x)

$$( 3 x - 4 ) ( 3 x + 4 ) + 5 ( 2 x + 1 ) ( 5 - 4 x )$$

Evaluate

$$9+30x-31x^{2}$$

Solution Steps

Consider \left(3x-4\right)\left(3x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(3x\right)^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Calculate 4 to the power of 2 and get 16.

$$9x^{2}-16+5\left(2x+1\right)\left(5-4x\right)$$

Use the distributive property to multiply 5 by 2x+1.

$$9x^{2}-16+\left(10x+5\right)\left(5-4x\right)$$

Apply the distributive property by multiplying each term of 10x+5 by each term of 5-4x.

$$9x^{2}-16+50x-40x^{2}+25-20x$$

Combine 50x and -20x to get 30x.

$$9x^{2}-16+30x-40x^{2}+25$$

Combine 9x^{2} and -40x^{2} to get -31x^{2}.

$$-31x^{2}-16+30x+25$$

Add -16 and 25 to get 9.

$$-31x^{2}+9+30x$$

Expand

$$9+30x-31x^{2}$$

Solution Steps

Consider \left(3x-4\right)\left(3x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(3x\right)^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4^{2}+5\left(2x+1\right)\left(5-4x\right)$$

Calculate 4 to the power of 2 and get 16.

$$9x^{2}-16+5\left(2x+1\right)\left(5-4x\right)$$

Use the distributive property to multiply 5 by 2x+1.

$$9x^{2}-16+\left(10x+5\right)\left(5-4x\right)$$

Apply the distributive property by multiplying each term of 10x+5 by each term of 5-4x.

$$9x^{2}-16+50x-40x^{2}+25-20x$$

Combine 50x and -20x to get 30x.

$$9x^{2}-16+30x-40x^{2}+25$$

Combine 9x^{2} and -40x^{2} to get -31x^{2}.

$$-31x^{2}-16+30x+25$$

Add -16 and 25 to get 9.

$$-31x^{2}+9+30x$$