Solve x(9x-5)-(3x+1)(3x-1)

$$x ( 9 x - 5 ) - ( 3 x + 1 ) ( 3 x - 1 )$$

Evaluate

$$1-5x$$

Solution Steps

Use the distributive property to multiply x by 9x-5.

$$9x^{2}-5x-\left(3x+1\right)\left(3x-1\right)$$

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}.

$$9x^{2}-5x-\left(\left(3x\right)^{2}-1^{2}\right)$$

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

$$9x^{2}-5x-\left(3^{2}x^{2}-1^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

The opposite of -1 is 1.

$$9x^{2}-5x-9x^{2}+1$$

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

$$-5x+1$$

Expand

$$1-5x$$

Solution Steps

Use the distributive property to multiply x by 9x-5.

$$9x^{2}-5x-\left(3x+1\right)\left(3x-1\right)$$

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}.

$$9x^{2}-5x-\left(\left(3x\right)^{2}-1^{2}\right)$$

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

$$9x^{2}-5x-\left(3^{2}x^{2}-1^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

The opposite of -1 is 1.

$$9x^{2}-5x-9x^{2}+1$$

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

$$-5x+1$$