Solve 2h^2-7h+6

$$2 { h }^{ 2 } -7h+6$$

Factor

$$\left(h-2\right)\left(2h-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 2h^{2}+ah+bh+6. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 6=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite 2h^{2}-7h+6 as \left(2h^{2}-4h\right)+\left(-3h+6\right).

$$\left(2h^{2}-4h\right)+\left(-3h+6\right)$$

Factor out 2h in the first and -3 in the second group.

$$2h\left(h-2\right)-3\left(h-2\right)$$

Factor out common term h-2 by using distributive property.

$$\left(h-2\right)\left(2h-3\right)$$

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.

$$2h^{2}-7h+6=0$$

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.

$$h=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}$$

Square -7.

$$h=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$h=\frac{-\left(-7\right)±\sqrt{49-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$h=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 2}$$

Add 49 to -48.

$$h=\frac{-\left(-7\right)±\sqrt{1}}{2\times 2}$$

Take the square root of 1.

$$h=\frac{-\left(-7\right)±1}{2\times 2}$$

The opposite of -7 is 7.

$$h=\frac{7±1}{2\times 2}$$

Multiply 2 times 2.

$$h=\frac{7±1}{4}$$

Now solve the equation h=\frac{7±1}{4} when ± is plus. Add 7 to 1.

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

Divide 8 by 4.

$$h=2$$

Now solve the equation h=\frac{7±1}{4} when ± is minus. Subtract 1 from 7.

$$h=\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$h=\frac{3}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and \frac{3}{2} for x_{2}.

$$2h^{2}-7h+6=2\left(h-2\right)\left(h-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from h by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$2h^{2}-7h+6=2\left(h-2\right)\times \frac{2h-3}{2}$$

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

$$2h^{2}-7h+6=\left(h-2\right)\left(2h-3\right)$$

Evaluate

$$\left(h-2\right)\left(2h-3\right)$$