Solve 2x^2-21x+52

$$2 { x }^{ 2 } -21x+52$$

Factor

$$\left(x-4\right)\left(2x-13\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+52. To find a and b, set up a system to be solved.

$$a+b=-21 ab=2\times 52=104$$

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 104.

$$-1,-104 -2,-52 -4,-26 -8,-13$$

Calculate the sum for each pair.

$$-1-104=-105 -2-52=-54 -4-26=-30 -8-13=-21$$

The solution is the pair that gives sum -21.

$$a=-13 b=-8$$

Rewrite 2x^{2}-21x+52 as \left(2x^{2}-13x\right)+\left(-8x+52\right).

$$\left(2x^{2}-13x\right)+\left(-8x+52\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(2x-13\right)-4\left(2x-13\right)$$

Factor out common term 2x-13 by using distributive property.

$$\left(2x-13\right)\left(x-4\right)$$

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.

$$2x^{2}-21x+52=0$$

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.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 2\times 52}}{2\times 2}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 2\times 52}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-21\right)±\sqrt{441-8\times 52}}{2\times 2}$$

Multiply -8 times 52.

$$x=\frac{-\left(-21\right)±\sqrt{441-416}}{2\times 2}$$

Add 441 to -416.

$$x=\frac{-\left(-21\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

$$x=\frac{-\left(-21\right)±5}{2\times 2}$$

The opposite of -21 is 21.

$$x=\frac{21±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{21±5}{4}$$

Now solve the equation x=\frac{21±5}{4} when ± is plus. Add 21 to 5.

$$x=\frac{26}{4}$$

Reduce the fraction \frac{26}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{2}$$

Now solve the equation x=\frac{21±5}{4} when ± is minus. Subtract 5 from 21.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

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

$$2x^{2}-21x+52=2\left(x-\frac{13}{2}\right)\left(x-4\right)$$

Subtract \frac{13}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2x^{2}-21x+52=2\times \frac{2x-13}{2}\left(x-4\right)$$

Cancel out 2, the greatest common factor in 2 and 2.

$$2x^{2}-21x+52=\left(2x-13\right)\left(x-4\right)$$

Evaluate

$$\left(x-4\right)\left(2x-13\right)$$