Solve (8x+1)^2+(3x-1)(3x+1)

$$( 8 x + 1 ) ^ { 2 } + ( 3 x - 1 ) ( 3 x + 1 )$$

Evaluate

$$x\left(73x+16\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8x+1\right)^{2}.

$$64x^{2}+16x+1+\left(3x-1\right)\left(3x+1\right)$$

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.

$$64x^{2}+16x+1+\left(3x\right)^{2}-1$$

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

$$64x^{2}+16x+1+3^{2}x^{2}-1$$

Calculate 3 to the power of 2 and get 9.

$$64x^{2}+16x+1+9x^{2}-1$$

Combine 64x^{2} and 9x^{2} to get 73x^{2}.

$$73x^{2}+16x+1-1$$

Subtract 1 from 1 to get 0.

$$73x^{2}+16x$$

Expand

$$73x^{2}+16x$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(8x+1\right)^{2}.

$$64x^{2}+16x+1+\left(3x-1\right)\left(3x+1\right)$$

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.

$$64x^{2}+16x+1+\left(3x\right)^{2}-1$$

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

$$64x^{2}+16x+1+3^{2}x^{2}-1$$

Calculate 3 to the power of 2 and get 9.

$$64x^{2}+16x+1+9x^{2}-1$$

Combine 64x^{2} and 9x^{2} to get 73x^{2}.

$$73x^{2}+16x+1-1$$

Subtract 1 from 1 to get 0.

$$73x^{2}+16x$$