Solve 6x^2-19x+10

$$6 { x }^{ 2 } -19x+10$$

Factor

$$\left(2x-5\right)\left(3x-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-19 ab=6\times 10=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -19.

$$a=-15 b=-4$$

Rewrite 6x^{2}-19x+10 as \left(6x^{2}-15x\right)+\left(-4x+10\right).

$$\left(6x^{2}-15x\right)+\left(-4x+10\right)$$

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

$$3x\left(2x-5\right)-2\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(3x-2\right)$$

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.

$$6x^{2}-19x+10=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 6\times 10}}{2\times 6}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 6\times 10}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-19\right)±\sqrt{361-24\times 10}}{2\times 6}$$

Multiply -24 times 10.

$$x=\frac{-\left(-19\right)±\sqrt{361-240}}{2\times 6}$$

Add 361 to -240.

$$x=\frac{-\left(-19\right)±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-\left(-19\right)±11}{2\times 6}$$

The opposite of -19 is 19.

$$x=\frac{19±11}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{19±11}{12}$$

Now solve the equation x=\frac{19±11}{12} when ± is plus. Add 19 to 11.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{19±11}{12} when ± is minus. Subtract 11 from 19.

$$x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and \frac{2}{3} for x_{2}.

$$6x^{2}-19x+10=6\left(x-\frac{5}{2}\right)\left(x-\frac{2}{3}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-19x+10=6\times \frac{2x-5}{2}\left(x-\frac{2}{3}\right)$$

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

$$6x^{2}-19x+10=6\times \frac{2x-5}{2}\times \frac{3x-2}{3}$$

Multiply \frac{2x-5}{2} times \frac{3x-2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-19x+10=6\times \frac{\left(2x-5\right)\left(3x-2\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-19x+10=6\times \frac{\left(2x-5\right)\left(3x-2\right)}{6}$$

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

$$6x^{2}-19x+10=\left(2x-5\right)\left(3x-2\right)$$

Evaluate

$$\left(2x-5\right)\left(3x-2\right)$$