Solve (5x^2-26x+5)

$$( 5 x ^ { 2 } - 26 x + 5 )$$

Factor

$$\left(x-5\right)\left(5x-1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-26 ab=5\times 5=25$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 25.

$$-1,-25 -5,-5$$

Calculate the sum for each pair.

$$-1-25=-26 -5-5=-10$$

The solution is the pair that gives sum -26.

$$a=-25 b=-1$$

Rewrite 5x^{2}-26x+5 as \left(5x^{2}-25x\right)+\left(-x+5\right).

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

Factor out 5x in the first and -1 in the second group.

$$5x\left(x-5\right)-\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(5x-1\right)$$

Steps Using the Quadratic Formula

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.

$$5x^{2}-26x+5=0$$

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.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 5\times 5}}{2\times 5}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 5\times 5}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-26\right)±\sqrt{676-20\times 5}}{2\times 5}$$

Multiply -20 times 5.

$$x=\frac{-\left(-26\right)±\sqrt{676-100}}{2\times 5}$$

Add 676 to -100.

$$x=\frac{-\left(-26\right)±\sqrt{576}}{2\times 5}$$

Take the square root of 576.

$$x=\frac{-\left(-26\right)±24}{2\times 5}$$

The opposite of -26 is 26.

$$x=\frac{26±24}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{26±24}{10}$$

Now solve the equation x=\frac{26±24}{10} when ± is plus. Add 26 to 24.

$$x=\frac{50}{10}$$

Divide 50 by 10.

$$x=5$$

Now solve the equation x=\frac{26±24}{10} when ± is minus. Subtract 24 from 26.

$$x=\frac{2}{10}$$

Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and \frac{1}{5} for x_{2}.

$$5x^{2}-26x+5=5\left(x-5\right)\left(x-\frac{1}{5}\right)$$

Subtract \frac{1}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-26x+5=5\left(x-5\right)\times \frac{5x-1}{5}$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}-26x+5=\left(x-5\right)\left(5x-1\right)$$

Evaluate

$$\left(x-5\right)\left(5x-1\right)$$