Solve (2x-3)(x-5)+(5-1)^0

$$( 2 x - 3 ) ( x - 5 ) + ( 5 - 1 ) ^ { 0 }$$

Evaluate

$$\left(x-5\right)\left(2x-3\right)+1$$

Solution Steps

Use the distributive property to multiply 2x-3 by x-5 and combine like terms.

$$2x^{2}-13x+15+\left(5-1\right)^{0}$$

Subtract 1 from 5 to get 4.

$$2x^{2}-13x+15+4^{0}$$

Calculate 4 to the power of 0 and get 1.

$$2x^{2}-13x+15+1$$

Add 15 and 1 to get 16.

$$2x^{2}-13x+16$$

Factor

$$2\left(x-\frac{13-\sqrt{41}}{4}\right)\left(x-\frac{\sqrt{41}+13}{4}\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-3 by x-5 and combine like terms.

$$factor(2x^{2}-13x+15+\left(5-1\right)^{0})$$

Subtract 1 from 5 to get 4.

$$factor(2x^{2}-13x+15+4^{0})$$

Calculate 4 to the power of 0 and get 1.

$$factor(2x^{2}-13x+15+1)$$

Add 15 and 1 to get 16.

$$factor(2x^{2}-13x+16)$$

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$2x^{2}-13x+16=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 2\times 16}}{2\times 2}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\times 2\times 16}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-13\right)±\sqrt{169-8\times 16}}{2\times 2}$$

Multiply -8 times 16.

$$x=\frac{-\left(-13\right)±\sqrt{169-128}}{2\times 2}$$

Add 169 to -128.

$$x=\frac{-\left(-13\right)±\sqrt{41}}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±\sqrt{41}}{4}$$

Now solve the equation x=\frac{13±\sqrt{41}}{4} when ± is plus. Add 13 to \sqrt{41}.

$$x=\frac{\sqrt{41}+13}{4}$$

Now solve the equation x=\frac{13±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from 13.

$$x=\frac{13-\sqrt{41}}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{13+\sqrt{41}}{4} for x_{1} and \frac{13-\sqrt{41}}{4} for x_{2}.

$$2x^{2}-13x+16=2\left(x-\frac{\sqrt{41}+13}{4}\right)\left(x-\frac{13-\sqrt{41}}{4}\right)$$