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

$$( 5 x - 4 ) ^ { 2 } - ( 5 x + 1 ) ( 5 x - 1 )$$

Evaluate

$$17-40x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(5x+1\right)\left(5x-1\right)$$

Consider \left(5x+1\right)\left(5x-1\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 1.

$$25x^{2}-40x+16-\left(\left(5x\right)^{2}-1\right)$$

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

$$25x^{2}-40x+16-\left(5^{2}x^{2}-1\right)$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-40x+16-\left(25x^{2}-1\right)$$

To find the opposite of 25x^{2}-1, find the opposite of each term.

$$25x^{2}-40x+16-25x^{2}+1$$

Combine 25x^{2} and -25x^{2} to get 0.

$$-40x+16+1$$

Add 16 and 1 to get 17.

$$-40x+17$$

Expand

$$17-40x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16-\left(5x+1\right)\left(5x-1\right)$$

Consider \left(5x+1\right)\left(5x-1\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 1.

$$25x^{2}-40x+16-\left(\left(5x\right)^{2}-1\right)$$

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

$$25x^{2}-40x+16-\left(5^{2}x^{2}-1\right)$$

Calculate 5 to the power of 2 and get 25.

$$25x^{2}-40x+16-\left(25x^{2}-1\right)$$

To find the opposite of 25x^{2}-1, find the opposite of each term.

$$25x^{2}-40x+16-25x^{2}+1$$

Combine 25x^{2} and -25x^{2} to get 0.

$$-40x+16+1$$

Add 16 and 1 to get 17.

$$-40x+17$$