Solve 5x^2+19x+12

$$5 { x }^{ 2 } +19x+12$$

Factor

$$\left(x+3\right)\left(5x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=19 ab=5\times 12=60$$

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 5x^{2}+19x+12 as \left(5x^{2}+4x\right)+\left(15x+12\right).

$$\left(5x^{2}+4x\right)+\left(15x+12\right)$$

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

$$x\left(5x+4\right)+3\left(5x+4\right)$$

Factor out common term 5x+4 by using distributive property.

$$\left(5x+4\right)\left(x+3\right)$$

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.

$$5x^{2}+19x+12=0$$

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.

$$x=\frac{-19±\sqrt{19^{2}-4\times 5\times 12}}{2\times 5}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 5\times 12}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-19±\sqrt{361-20\times 12}}{2\times 5}$$

Multiply -20 times 12.

$$x=\frac{-19±\sqrt{361-240}}{2\times 5}$$

Add 361 to -240.

$$x=\frac{-19±\sqrt{121}}{2\times 5}$$

Take the square root of 121.

$$x=\frac{-19±11}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-19±11}{10}$$

Now solve the equation x=\frac{-19±11}{10} when ± is plus. Add -19 to 11.

$$x=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{-19±11}{10} when ± is minus. Subtract 11 from -19.

$$x=-\frac{30}{10}$$

Divide -30 by 10.

$$x=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{4}{5} for x_{1} and -3 for x_{2}.

$$5x^{2}+19x+12=5\left(x-\left(-\frac{4}{5}\right)\right)\left(x-\left(-3\right)\right)$$

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

$$5x^{2}+19x+12=5\left(x+\frac{4}{5}\right)\left(x+3\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+19x+12=5\times \frac{5x+4}{5}\left(x+3\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5x^{2}+19x+12=\left(5x+4\right)\left(x+3\right)$$

Evaluate

$$\left(x+3\right)\left(5x+4\right)$$