Solve 5x^2+13x+6

$$5 { x }^{ 2 } +13x+6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=13 ab=5\times 6=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=3 b=10$$

Rewrite 5x^{2}+13x+6 as \left(5x^{2}+3x\right)+\left(10x+6\right).

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

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

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

Factor out common term 5x+3 by using distributive property.

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

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.

$$5x^{2}+13x+6=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 5\times 6}}{2\times 5}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-13±\sqrt{169-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$x=\frac{-13±\sqrt{169-120}}{2\times 5}$$

Add 169 to -120.

$$x=\frac{-13±\sqrt{49}}{2\times 5}$$

Take the square root of 49.

$$x=\frac{-13±7}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-13±7}{10}$$

Now solve the equation x=\frac{-13±7}{10} when ± is plus. Add -13 to 7.

$$x=-\frac{6}{10}$$

Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{5}$$

Now solve the equation x=\frac{-13±7}{10} when ± is minus. Subtract 7 from -13.

$$x=-\frac{20}{10}$$

Divide -20 by 10.

$$x=-2$$

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

$$5x^{2}+13x+6=5\left(x-\left(-\frac{3}{5}\right)\right)\left(x-\left(-2\right)\right)$$

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

$$5x^{2}+13x+6=5\left(x+\frac{3}{5}\right)\left(x+2\right)$$

Add \frac{3}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+13x+6=5\times \frac{5x+3}{5}\left(x+2\right)$$

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

$$5x^{2}+13x+6=\left(5x+3\right)\left(x+2\right)$$

Evaluate

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