Solve 45x^2+250x-120

$$45 { x }^{ 2 } +250x-120$$

Factor

$$5\left(9x-4\right)\left(x+6\right)$$

Solution Steps

Factor out 5.

$$5\left(9x^{2}+50x-24\right)$$

Consider 9x^{2}+50x-24. Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=50 ab=9\left(-24\right)=-216$$

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 -216.

$$-1,216 -2,108 -3,72 -4,54 -6,36 -8,27 -9,24 -12,18$$

Calculate the sum for each pair.

$$-1+216=215 -2+108=106 -3+72=69 -4+54=50 -6+36=30 -8+27=19 -9+24=15 -12+18=6$$

The solution is the pair that gives sum 50.

$$a=-4 b=54$$

Rewrite 9x^{2}+50x-24 as \left(9x^{2}-4x\right)+\left(54x-24\right).

$$\left(9x^{2}-4x\right)+\left(54x-24\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(9x-4\right)+6\left(9x-4\right)$$

Factor out common term 9x-4 by using distributive property.

$$\left(9x-4\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$5\left(9x-4\right)\left(x+6\right)$$

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.

$$45x^{2}+250x-120=0$$

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.

$$x=\frac{-250±\sqrt{250^{2}-4\times 45\left(-120\right)}}{2\times 45}$$

Square 250.

$$x=\frac{-250±\sqrt{62500-4\times 45\left(-120\right)}}{2\times 45}$$

Multiply -4 times 45.

$$x=\frac{-250±\sqrt{62500-180\left(-120\right)}}{2\times 45}$$

Multiply -180 times -120.

$$x=\frac{-250±\sqrt{62500+21600}}{2\times 45}$$

Add 62500 to 21600.

$$x=\frac{-250±\sqrt{84100}}{2\times 45}$$

Take the square root of 84100.

$$x=\frac{-250±290}{2\times 45}$$

Multiply 2 times 45.

$$x=\frac{-250±290}{90}$$

Now solve the equation x=\frac{-250±290}{90} when ± is plus. Add -250 to 290.

$$x=\frac{40}{90}$$

Reduce the fraction \frac{40}{90} to lowest terms by extracting and canceling out 10.

$$x=\frac{4}{9}$$

Now solve the equation x=\frac{-250±290}{90} when ± is minus. Subtract 290 from -250.

$$x=-\frac{540}{90}$$

Divide -540 by 90.

$$x=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{9} for x_{1} and -6 for x_{2}.

$$45x^{2}+250x-120=45\left(x-\frac{4}{9}\right)\left(x-\left(-6\right)\right)$$

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

$$45x^{2}+250x-120=45\left(x-\frac{4}{9}\right)\left(x+6\right)$$

Subtract \frac{4}{9} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$45x^{2}+250x-120=45\times \frac{9x-4}{9}\left(x+6\right)$$

Cancel out 9, the greatest common factor in 45 and 9.

$$45x^{2}+250x-120=5\left(9x-4\right)\left(x+6\right)$$

Evaluate

$$5\left(9x-4\right)\left(x+6\right)$$