Solve (3x-7)^2-(3x+1)(3x-1)

$$( 3 x - 7 ) ^ { 2 } - ( 3 x + 1 ) ( 3 x - 1 )$$

Evaluate

$$50-42x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-\left(3x+1\right)\left(3x-1\right)$$

Consider \left(3x+1\right)\left(3x-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.

$$9x^{2}-42x+49-\left(\left(3x\right)^{2}-1\right)$$

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

$$9x^{2}-42x+49-\left(3^{2}x^{2}-1\right)$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-42x+49-\left(9x^{2}-1\right)$$

To find the opposite of 9x^{2}-1, find the opposite of each term.

$$9x^{2}-42x+49-9x^{2}+1$$

Combine 9x^{2} and -9x^{2} to get 0.

$$-42x+49+1$$

Add 49 and 1 to get 50.

$$-42x+50$$

Expand

$$50-42x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-7\right)^{2}.

$$9x^{2}-42x+49-\left(3x+1\right)\left(3x-1\right)$$

Consider \left(3x+1\right)\left(3x-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.

$$9x^{2}-42x+49-\left(\left(3x\right)^{2}-1\right)$$

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

$$9x^{2}-42x+49-\left(3^{2}x^{2}-1\right)$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-42x+49-\left(9x^{2}-1\right)$$

To find the opposite of 9x^{2}-1, find the opposite of each term.

$$9x^{2}-42x+49-9x^{2}+1$$

Combine 9x^{2} and -9x^{2} to get 0.

$$-42x+49+1$$

Add 49 and 1 to get 50.

$$-42x+50$$