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

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

Evaluate

$$48x^{2}+6x-13$$

Solution Steps

Consider \left(7x-2\right)\left(7x+2\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 2.

$$\left(7x\right)^{2}-4-\left(x-3\right)^{2}$$

Expand \left(7x\right)^{2}.

$$7^{2}x^{2}-4-\left(x-3\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49x^{2}-4-\left(x-3\right)^{2}$$

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

$$49x^{2}-4-\left(x^{2}-6x+9\right)$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$49x^{2}-4-x^{2}+6x-9$$

Combine 49x^{2} and -x^{2} to get 48x^{2}.

$$48x^{2}-4+6x-9$$

Subtract 9 from -4 to get -13.

$$48x^{2}-13+6x$$

Expand

$$48x^{2}+6x-13$$

Solution Steps

Consider \left(7x-2\right)\left(7x+2\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 2.

$$\left(7x\right)^{2}-4-\left(x-3\right)^{2}$$

Expand \left(7x\right)^{2}.

$$7^{2}x^{2}-4-\left(x-3\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49x^{2}-4-\left(x-3\right)^{2}$$

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

$$49x^{2}-4-\left(x^{2}-6x+9\right)$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$49x^{2}-4-x^{2}+6x-9$$

Combine 49x^{2} and -x^{2} to get 48x^{2}.

$$48x^{2}-4+6x-9$$

Subtract 9 from -4 to get -13.

$$48x^{2}-13+6x$$