Solve 30x^2+11x-30

$$30 { x }^{ 2 } +11x-30$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 30x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=11 ab=30\left(-30\right)=-900$$

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 -900.

$$-1,900 -2,450 -3,300 -4,225 -5,180 -6,150 -9,100 -10,90 -12,75 -15,60 -18,50 -20,45 -25,36 -30,30$$

Calculate the sum for each pair.

$$-1+900=899 -2+450=448 -3+300=297 -4+225=221 -5+180=175 -6+150=144 -9+100=91 -10+90=80 -12+75=63 -15+60=45 -18+50=32 -20+45=25 -25+36=11 -30+30=0$$

The solution is the pair that gives sum 11.

$$a=-25 b=36$$

Rewrite 30x^{2}+11x-30 as \left(30x^{2}-25x\right)+\left(36x-30\right).

$$\left(30x^{2}-25x\right)+\left(36x-30\right)$$

Factor out 5x in the first and 6 in the second group.

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

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

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

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.

$$30x^{2}+11x-30=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\times 30\left(-30\right)}}{2\times 30}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 30\left(-30\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-11±\sqrt{121-120\left(-30\right)}}{2\times 30}$$

Multiply -120 times -30.

$$x=\frac{-11±\sqrt{121+3600}}{2\times 30}$$

Add 121 to 3600.

$$x=\frac{-11±\sqrt{3721}}{2\times 30}$$

Take the square root of 3721.

$$x=\frac{-11±61}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{-11±61}{60}$$

Now solve the equation x=\frac{-11±61}{60} when ± is plus. Add -11 to 61.

$$x=\frac{50}{60}$$

Reduce the fraction \frac{50}{60} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{-11±61}{60} when ± is minus. Subtract 61 from -11.

$$x=-\frac{72}{60}$$

Reduce the fraction \frac{-72}{60} to lowest terms by extracting and canceling out 12.

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

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

$$30x^{2}+11x-30=30\left(x-\frac{5}{6}\right)\left(x-\left(-\frac{6}{5}\right)\right)$$

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

$$30x^{2}+11x-30=30\left(x-\frac{5}{6}\right)\left(x+\frac{6}{5}\right)$$

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

$$30x^{2}+11x-30=30\times \frac{6x-5}{6}\left(x+\frac{6}{5}\right)$$

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

$$30x^{2}+11x-30=30\times \frac{6x-5}{6}\times \frac{5x+6}{5}$$

Multiply \frac{6x-5}{6} times \frac{5x+6}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+11x-30=30\times \frac{\left(6x-5\right)\left(5x+6\right)}{6\times 5}$$

Multiply 6 times 5.

$$30x^{2}+11x-30=30\times \frac{\left(6x-5\right)\left(5x+6\right)}{30}$$

Cancel out 30, the greatest common factor in 30 and 30.

$$30x^{2}+11x-30=\left(6x-5\right)\left(5x+6\right)$$

Evaluate

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