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

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

Evaluate

$$-x-16$$

Solution Steps

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

$$x^{2}-6x+9+\left(2x+5\right)\left(2x-5\right)-5x\left(x-1\right)$$

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.

$$x^{2}-6x+9+\left(2x\right)^{2}-25-5x\left(x-1\right)$$

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

$$x^{2}-6x+9+2^{2}x^{2}-25-5x\left(x-1\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-6x+9+4x^{2}-25-5x\left(x-1\right)$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-6x+9-25-5x\left(x-1\right)$$

Subtract 25 from 9 to get -16.

$$5x^{2}-6x-16-5x\left(x-1\right)$$

Use the distributive property to multiply -5x by x-1.

$$5x^{2}-6x-16-5x^{2}+5x$$

Combine 5x^{2} and -5x^{2} to get 0.

$$-6x-16+5x$$

Combine -6x and 5x to get -x.

$$-x-16$$

Expand

$$-x-16$$

Solution Steps

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

$$x^{2}-6x+9+\left(2x+5\right)\left(2x-5\right)-5x\left(x-1\right)$$

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.

$$x^{2}-6x+9+\left(2x\right)^{2}-25-5x\left(x-1\right)$$

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

$$x^{2}-6x+9+2^{2}x^{2}-25-5x\left(x-1\right)$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-6x+9+4x^{2}-25-5x\left(x-1\right)$$

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}-6x+9-25-5x\left(x-1\right)$$

Subtract 25 from 9 to get -16.

$$5x^{2}-6x-16-5x\left(x-1\right)$$

Use the distributive property to multiply -5x by x-1.

$$5x^{2}-6x-16-5x^{2}+5x$$

Combine 5x^{2} and -5x^{2} to get 0.

$$-6x-16+5x$$

Combine -6x and 5x to get -x.

$$-x-16$$