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

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

Evaluate

$$12x^{2}+28x-98$$

Solution Steps

Consider \left(4x+7\right)\left(4x-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.

$$\left(4x\right)^{2}-49-\left(2x-7\right)^{2}$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-49-\left(2x-7\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-49-\left(2x-7\right)^{2}$$

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

$$16x^{2}-49-\left(4x^{2}-28x+49\right)$$

To find the opposite of 4x^{2}-28x+49, find the opposite of each term.

$$16x^{2}-49-4x^{2}+28x-49$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}-49+28x-49$$

Subtract 49 from -49 to get -98.

$$12x^{2}-98+28x$$

Expand

$$12x^{2}+28x-98$$

Solution Steps

Consider \left(4x+7\right)\left(4x-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.

$$\left(4x\right)^{2}-49-\left(2x-7\right)^{2}$$

Expand \left(4x\right)^{2}.

$$4^{2}x^{2}-49-\left(2x-7\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-49-\left(2x-7\right)^{2}$$

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

$$16x^{2}-49-\left(4x^{2}-28x+49\right)$$

To find the opposite of 4x^{2}-28x+49, find the opposite of each term.

$$16x^{2}-49-4x^{2}+28x-49$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}-49+28x-49$$

Subtract 49 from -49 to get -98.

$$12x^{2}-98+28x$$