Solve 5x^2+9x-44

$$5 { x }^{ 2 } +9x-44$$

Factor

$$\left(5x-11\right)\left(x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-44. To find a and b, set up a system to be solved.

$$a+b=9 ab=5\left(-44\right)=-220$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -220.

$$-1,220 -2,110 -4,55 -5,44 -10,22 -11,20$$

Calculate the sum for each pair.

$$-1+220=219 -2+110=108 -4+55=51 -5+44=39 -10+22=12 -11+20=9$$

The solution is the pair that gives sum 9.

$$a=-11 b=20$$

Rewrite 5x^{2}+9x-44 as \left(5x^{2}-11x\right)+\left(20x-44\right).

$$\left(5x^{2}-11x\right)+\left(20x-44\right)$$

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

$$x\left(5x-11\right)+4\left(5x-11\right)$$

Factor out common term 5x-11 by using distributive property.

$$\left(5x-11\right)\left(x+4\right)$$

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.

$$5x^{2}+9x-44=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 5\left(-44\right)}}{2\times 5}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 5\left(-44\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-9±\sqrt{81-20\left(-44\right)}}{2\times 5}$$

Multiply -20 times -44.

$$x=\frac{-9±\sqrt{81+880}}{2\times 5}$$

Add 81 to 880.

$$x=\frac{-9±\sqrt{961}}{2\times 5}$$

Take the square root of 961.

$$x=\frac{-9±31}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-9±31}{10}$$

Now solve the equation x=\frac{-9±31}{10} when ± is plus. Add -9 to 31.

$$x=\frac{22}{10}$$

Reduce the fraction \frac{22}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{5}$$

Now solve the equation x=\frac{-9±31}{10} when ± is minus. Subtract 31 from -9.

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

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{11}{5} for x_{1} and -4 for x_{2}.

$$5x^{2}+9x-44=5\left(x-\frac{11}{5}\right)\left(x-\left(-4\right)\right)$$

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

$$5x^{2}+9x-44=5\left(x-\frac{11}{5}\right)\left(x+4\right)$$

Subtract \frac{11}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}+9x-44=5\times \frac{5x-11}{5}\left(x+4\right)$$

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

$$5x^{2}+9x-44=\left(5x-11\right)\left(x+4\right)$$

Evaluate

$$\left(5x-11\right)\left(x+4\right)$$