Solve 26x^2-23x+5

$$26 { x }^{ 2 } -23x+5$$

Factor

$$\left(2x-1\right)\left(13x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 26x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-23 ab=26\times 5=130$$

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 130.

$$-1,-130 -2,-65 -5,-26 -10,-13$$

Calculate the sum for each pair.

$$-1-130=-131 -2-65=-67 -5-26=-31 -10-13=-23$$

The solution is the pair that gives sum -23.

$$a=-13 b=-10$$

Rewrite 26x^{2}-23x+5 as \left(26x^{2}-13x\right)+\left(-10x+5\right).

$$\left(26x^{2}-13x\right)+\left(-10x+5\right)$$

Factor out 13x in the first and -5 in the second group.

$$13x\left(2x-1\right)-5\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(13x-5\right)$$

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.

$$26x^{2}-23x+5=0$$

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.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 26\times 5}}{2\times 26}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 26\times 5}}{2\times 26}$$

Multiply -4 times 26.

$$x=\frac{-\left(-23\right)±\sqrt{529-104\times 5}}{2\times 26}$$

Multiply -104 times 5.

$$x=\frac{-\left(-23\right)±\sqrt{529-520}}{2\times 26}$$

Add 529 to -520.

$$x=\frac{-\left(-23\right)±\sqrt{9}}{2\times 26}$$

Take the square root of 9.

$$x=\frac{-\left(-23\right)±3}{2\times 26}$$

The opposite of -23 is 23.

$$x=\frac{23±3}{2\times 26}$$

Multiply 2 times 26.

$$x=\frac{23±3}{52}$$

Now solve the equation x=\frac{23±3}{52} when ± is plus. Add 23 to 3.

$$x=\frac{26}{52}$$

Reduce the fraction \frac{26}{52} to lowest terms by extracting and canceling out 26.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{23±3}{52} when ± is minus. Subtract 3 from 23.

$$x=\frac{20}{52}$$

Reduce the fraction \frac{20}{52} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{13}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and \frac{5}{13} for x_{2}.

$$26x^{2}-23x+5=26\left(x-\frac{1}{2}\right)\left(x-\frac{5}{13}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$26x^{2}-23x+5=26\times \frac{2x-1}{2}\left(x-\frac{5}{13}\right)$$

Subtract \frac{5}{13} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$26x^{2}-23x+5=26\times \frac{2x-1}{2}\times \frac{13x-5}{13}$$

Multiply \frac{2x-1}{2} times \frac{13x-5}{13} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$26x^{2}-23x+5=26\times \frac{\left(2x-1\right)\left(13x-5\right)}{2\times 13}$$

Multiply 2 times 13.

$$26x^{2}-23x+5=26\times \frac{\left(2x-1\right)\left(13x-5\right)}{26}$$

Cancel out 26, the greatest common factor in 26 and 26.

$$26x^{2}-23x+5=\left(2x-1\right)\left(13x-5\right)$$

Evaluate

$$\left(2x-1\right)\left(13x-5\right)$$