Solve 19x-6x^2-14

$$19 x - 6 x ^ { 2 } - 14$$

Factor

$$\left(2-x\right)\left(6x-7\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-6x^{2}+19x-14$$

Factor the expression by grouping. First, the expression needs to be rewritten as -6x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=19 ab=-6\left(-14\right)=84$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 84.

$$1,84 2,42 3,28 4,21 6,14 7,12$$

Calculate the sum for each pair.

$$1+84=85 2+42=44 3+28=31 4+21=25 6+14=20 7+12=19$$

The solution is the pair that gives sum 19.

$$a=12 b=7$$

Rewrite -6x^{2}+19x-14 as \left(-6x^{2}+12x\right)+\left(7x-14\right).

$$\left(-6x^{2}+12x\right)+\left(7x-14\right)$$

Factor out 6x in the first and -7 in the second group.

$$6x\left(-x+2\right)-7\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(6x-7\right)$$

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.

$$-6x^{2}+19x-14=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\left(-6\right)\left(-14\right)}}{2\left(-6\right)}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\left(-6\right)\left(-14\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-19±\sqrt{361+24\left(-14\right)}}{2\left(-6\right)}$$

Multiply 24 times -14.

$$x=\frac{-19±\sqrt{361-336}}{2\left(-6\right)}$$

Add 361 to -336.

$$x=\frac{-19±\sqrt{25}}{2\left(-6\right)}$$

Take the square root of 25.

$$x=\frac{-19±5}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-19±5}{-12}$$

Now solve the equation x=\frac{-19±5}{-12} when ± is plus. Add -19 to 5.

$$x=-\frac{14}{-12}$$

Reduce the fraction \frac{-14}{-12} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{6}$$

Now solve the equation x=\frac{-19±5}{-12} when ± is minus. Subtract 5 from -19.

$$x=-\frac{24}{-12}$$

Divide -24 by -12.

$$x=2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{6} for x_{1} and 2 for x_{2}.

$$-6x^{2}+19x-14=-6\left(x-\frac{7}{6}\right)\left(x-2\right)$$

Subtract \frac{7}{6} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}+19x-14=-6\times \frac{-6x+7}{-6}\left(x-2\right)$$

Cancel out 6, the greatest common factor in -6 and 6.

$$-6x^{2}+19x-14=\left(-6x+7\right)\left(x-2\right)$$

Evaluate

$$\left(2-x\right)\left(6x-7\right)$$