Solve 2x(3x-1)+(4x-3)^2+(2x+5)(2x-5)+(x+2)^3=

$$2 x ( 3 x - 1 ) + ( 4 x - 3 ) ^ { 2 } + ( 2 x + 5 ) ( 2 x - 5 ) + ( x + 2 ) ^ { 3 } =$$

Evaluate

$$x^{3}+32x^{2}-14x-8$$

Solution Steps

Use the distributive property to multiply 2x by 3x-1.

$$6x^{2}-2x+\left(4x-3\right)^{2}+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

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

$$6x^{2}-2x+16x^{2}-24x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Combine 6x^{2} and 16x^{2} to get 22x^{2}.

$$22x^{2}-2x-24x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Combine -2x and -24x to get -26x.

$$22x^{2}-26x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Consider \left(2x+5\right)\left(2x-5\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.

$$22x^{2}-26x+9+\left(2x\right)^{2}-25+\left(x+2\right)^{3}$$

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

$$22x^{2}-26x+9+2^{2}x^{2}-25+\left(x+2\right)^{3}$$

Calculate 2 to the power of 2 and get 4.

$$22x^{2}-26x+9+4x^{2}-25+\left(x+2\right)^{3}$$

Combine 22x^{2} and 4x^{2} to get 26x^{2}.

$$26x^{2}-26x+9-25+\left(x+2\right)^{3}$$

Subtract 25 from 9 to get -16.

$$26x^{2}-26x-16+\left(x+2\right)^{3}$$

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

$$26x^{2}-26x-16+x^{3}+6x^{2}+12x+8$$

Combine 26x^{2} and 6x^{2} to get 32x^{2}.

$$32x^{2}-26x-16+x^{3}+12x+8$$

Combine -26x and 12x to get -14x.

$$32x^{2}-14x-16+x^{3}+8$$

Add -16 and 8 to get -8.

$$32x^{2}-14x-8+x^{3}$$

Expand

$$x^{3}+32x^{2}-14x-8$$

Solution Steps

Use the distributive property to multiply 2x by 3x-1.

$$6x^{2}-2x+\left(4x-3\right)^{2}+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

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

$$6x^{2}-2x+16x^{2}-24x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Combine 6x^{2} and 16x^{2} to get 22x^{2}.

$$22x^{2}-2x-24x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Combine -2x and -24x to get -26x.

$$22x^{2}-26x+9+\left(2x+5\right)\left(2x-5\right)+\left(x+2\right)^{3}$$

Consider \left(2x+5\right)\left(2x-5\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.

$$22x^{2}-26x+9+\left(2x\right)^{2}-25+\left(x+2\right)^{3}$$

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

$$22x^{2}-26x+9+2^{2}x^{2}-25+\left(x+2\right)^{3}$$

Calculate 2 to the power of 2 and get 4.

$$22x^{2}-26x+9+4x^{2}-25+\left(x+2\right)^{3}$$

Combine 22x^{2} and 4x^{2} to get 26x^{2}.

$$26x^{2}-26x+9-25+\left(x+2\right)^{3}$$

Subtract 25 from 9 to get -16.

$$26x^{2}-26x-16+\left(x+2\right)^{3}$$

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

$$26x^{2}-26x-16+x^{3}+6x^{2}+12x+8$$

Combine 26x^{2} and 6x^{2} to get 32x^{2}.

$$32x^{2}-26x-16+x^{3}+12x+8$$

Combine -26x and 12x to get -14x.

$$32x^{2}-14x-16+x^{3}+8$$

Add -16 and 8 to get -8.

$$32x^{2}-14x-8+x^{3}$$