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 positive, a and b are both positive. 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.

$$a=4 b=15$$

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

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

Factor out 2x in the first and 5 in the second group.

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

Factor out common term 3x+2 by using distributive property.

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

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.

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

Square 19.

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

Multiply -4 times 6.

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

Multiply -24 times 10.

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

Add 361 to -240.

$$x=\frac{-19±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$x=\frac{-19±11}{2\times 6}$$

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.

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

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

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

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

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

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

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

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

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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

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

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

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

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

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

Multiply 3 times 2.

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

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

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

Evaluate

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