Solve (4x-7)^2+(5x-7)(5x+7)+96x

$$( 4 x - 7 ) ^ { 2 } + ( 5 x - 7 ) ( 5 x + 7 ) + 96 x$$

Evaluate

$$x\left(41x+40\right)$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-7\right)^{2}.

$$16x^{2}-56x+49+\left(5x-7\right)\left(5x+7\right)+96x$$

Consider \left(5x-7\right)\left(5x+7\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 7.

$$16x^{2}-56x+49+\left(5x\right)^{2}-49+96x$$

Expand \left(5x\right)^{2}.

$$16x^{2}-56x+49+5^{2}x^{2}-49+96x$$

Calculate 5 to the power of 2 and get 25.

$$16x^{2}-56x+49+25x^{2}-49+96x$$

Combine 16x^{2} and 25x^{2} to get 41x^{2}.

$$41x^{2}-56x+49-49+96x$$

Subtract 49 from 49 to get 0.

$$41x^{2}-56x+96x$$

Combine -56x and 96x to get 40x.

$$41x^{2}+40x$$

Expand

$$41x^{2}+40x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-7\right)^{2}.

$$16x^{2}-56x+49+\left(5x-7\right)\left(5x+7\right)+96x$$

Consider \left(5x-7\right)\left(5x+7\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 7.

$$16x^{2}-56x+49+\left(5x\right)^{2}-49+96x$$

Expand \left(5x\right)^{2}.

$$16x^{2}-56x+49+5^{2}x^{2}-49+96x$$

Calculate 5 to the power of 2 and get 25.

$$16x^{2}-56x+49+25x^{2}-49+96x$$

Combine 16x^{2} and 25x^{2} to get 41x^{2}.

$$41x^{2}-56x+49-49+96x$$

Subtract 49 from 49 to get 0.

$$41x^{2}-56x+96x$$

Combine -56x and 96x to get 40x.

$$41x^{2}+40x$$