Solve 5x^2+26x+24

$$5 { x }^{ 2 } +26x+24$$

Factor

$$\left(x+4\right)\left(5x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=26 ab=5\times 24=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 26.

$$a=6 b=20$$

Rewrite 5x^{2}+26x+24 as \left(5x^{2}+6x\right)+\left(20x+24\right).

$$\left(5x^{2}+6x\right)+\left(20x+24\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(5x+6\right)+4\left(5x+6\right)$$

Factor out common term 5x+6 by using distributive property.

$$\left(5x+6\right)\left(x+4\right)$$

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.

$$5x^{2}+26x+24=0$$

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.

$$x=\frac{-26±\sqrt{26^{2}-4\times 5\times 24}}{2\times 5}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\times 5\times 24}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-26±\sqrt{676-20\times 24}}{2\times 5}$$

Multiply -20 times 24.

$$x=\frac{-26±\sqrt{676-480}}{2\times 5}$$

Add 676 to -480.

$$x=\frac{-26±\sqrt{196}}{2\times 5}$$

Take the square root of 196.

$$x=\frac{-26±14}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-26±14}{10}$$

Now solve the equation x=\frac{-26±14}{10} when ± is plus. Add -26 to 14.

$$x=-\frac{12}{10}$$

Reduce the fraction \frac{-12}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{6}{5}$$

Now solve the equation x=\frac{-26±14}{10} when ± is minus. Subtract 14 from -26.

$$x=-\frac{40}{10}$$

Divide -40 by 10.

$$x=-4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{6}{5} for x_{1} and -4 for x_{2}.

$$5x^{2}+26x+24=5\left(x-\left(-\frac{6}{5}\right)\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+26x+24=5\left(x+\frac{6}{5}\right)\left(x+4\right)$$

Add \frac{6}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+26x+24=5\times \frac{5x+6}{5}\left(x+4\right)$$

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

$$5x^{2}+26x+24=\left(5x+6\right)\left(x+4\right)$$

Evaluate

$$\left(x+4\right)\left(5x+6\right)$$