Solve 2y^2-11y+14

$$2 { y }^{ 2 } -11y+14$$

Factor

$$\left(y-2\right)\left(2y-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2y^{2}+ay+by+14. To find a and b, set up a system to be solved.

$$a+b=-11 ab=2\times 14=28$$

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 28.

$$-1,-28 -2,-14 -4,-7$$

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -11.

$$a=-7 b=-4$$

Rewrite 2y^{2}-11y+14 as \left(2y^{2}-7y\right)+\left(-4y+14\right).

$$\left(2y^{2}-7y\right)+\left(-4y+14\right)$$

Factor out y in the first and -2 in the second group.

$$y\left(2y-7\right)-2\left(2y-7\right)$$

Factor out common term 2y-7 by using distributive property.

$$\left(2y-7\right)\left(y-2\right)$$

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.

$$2y^{2}-11y+14=0$$

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.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\times 14}}{2\times 2}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times 2\times 14}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-11\right)±\sqrt{121-8\times 14}}{2\times 2}$$

Multiply -8 times 14.

$$y=\frac{-\left(-11\right)±\sqrt{121-112}}{2\times 2}$$

Add 121 to -112.

$$y=\frac{-\left(-11\right)±\sqrt{9}}{2\times 2}$$

Take the square root of 9.

$$y=\frac{-\left(-11\right)±3}{2\times 2}$$

The opposite of -11 is 11.

$$y=\frac{11±3}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{11±3}{4}$$

Now solve the equation y=\frac{11±3}{4} when ± is plus. Add 11 to 3.

$$y=\frac{14}{4}$$

Reduce the fraction \frac{14}{4} to lowest terms by extracting and canceling out 2.

$$y=\frac{7}{2}$$

Now solve the equation y=\frac{11±3}{4} when ± is minus. Subtract 3 from 11.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and 2 for x_{2}.

$$2y^{2}-11y+14=2\left(y-\frac{7}{2}\right)\left(y-2\right)$$

Subtract \frac{7}{2} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2y^{2}-11y+14=2\times \frac{2y-7}{2}\left(y-2\right)$$

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

$$2y^{2}-11y+14=\left(2y-7\right)\left(y-2\right)$$

Evaluate

$$\left(y-2\right)\left(2y-7\right)$$