Solve 30x^2+5x-10

$$30 { x }^{ 2 } +5x-10$$

Factor

$$5\left(2x-1\right)\left(3x+2\right)$$

Solution Steps

Factor out 5.

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

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

$$a+b=1 ab=6\left(-2\right)=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite 6x^{2}+x-2 as \left(6x^{2}-3x\right)+\left(4x-2\right).

$$\left(6x^{2}-3x\right)+\left(4x-2\right)$$

Factor out 3x in the first and 2 in the second group.

$$3x\left(2x-1\right)+2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(3x+2\right)$$

Rewrite the complete factored expression.

$$5\left(2x-1\right)\left(3x+2\right)$$

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.

$$30x^{2}+5x-10=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\times 30\left(-10\right)}}{2\times 30}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 30\left(-10\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-5±\sqrt{25-120\left(-10\right)}}{2\times 30}$$

Multiply -120 times -10.

$$x=\frac{-5±\sqrt{25+1200}}{2\times 30}$$

Add 25 to 1200.

$$x=\frac{-5±\sqrt{1225}}{2\times 30}$$

Take the square root of 1225.

$$x=\frac{-5±35}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{-5±35}{60}$$

Now solve the equation x=\frac{-5±35}{60} when ± is plus. Add -5 to 35.

$$x=\frac{30}{60}$$

Reduce the fraction \frac{30}{60} to lowest terms by extracting and canceling out 30.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-5±35}{60} when ± is minus. Subtract 35 from -5.

$$x=-\frac{40}{60}$$

Reduce the fraction \frac{-40}{60} to lowest terms by extracting and canceling out 20.

$$x=-\frac{2}{3}$$

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

$$30x^{2}+5x-10=30\left(x-\frac{1}{2}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$30x^{2}+5x-10=30\left(x-\frac{1}{2}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+5x-10=30\times \frac{2x-1}{2}\left(x+\frac{2}{3}\right)$$

Add \frac{2}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+5x-10=30\times \frac{2x-1}{2}\times \frac{3x+2}{3}$$

Multiply \frac{2x-1}{2} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30x^{2}+5x-10=30\times \frac{\left(2x-1\right)\left(3x+2\right)}{2\times 3}$$

Multiply 2 times 3.

$$30x^{2}+5x-10=30\times \frac{\left(2x-1\right)\left(3x+2\right)}{6}$$

Cancel out 6, the greatest common factor in 30 and 6.

$$30x^{2}+5x-10=5\left(2x-1\right)\left(3x+2\right)$$

Evaluate

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