Solve (3x-4)^2+(5+3x)(5-3x)

$$( 3 x - 4 ) ^ { 2 } + ( 5 + 3 x ) ( 5 - 3 x )$$

Evaluate

$$41-24x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16+\left(5+3x\right)\left(5-3x\right)$$

Consider \left(5+3x\right)\left(5-3x\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 5.

$$9x^{2}-24x+16+25-\left(3x\right)^{2}$$

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

$$9x^{2}-24x+16+25-3^{2}x^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-24x+16+25-9x^{2}$$

Add 16 and 25 to get 41.

$$9x^{2}-24x+41-9x^{2}$$

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

$$-24x+41$$

Expand

$$41-24x$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$9x^{2}-24x+16+\left(5+3x\right)\left(5-3x\right)$$

Consider \left(5+3x\right)\left(5-3x\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 5.

$$9x^{2}-24x+16+25-\left(3x\right)^{2}$$

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

$$9x^{2}-24x+16+25-3^{2}x^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-24x+16+25-9x^{2}$$

Add 16 and 25 to get 41.

$$9x^{2}-24x+41-9x^{2}$$

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

$$-24x+41$$