Solve 6x^2-7x-20

$$6 { x }^{ 2 } -7x-20$$

Factor

$$\left(2x-5\right)\left(3x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=-7 ab=6\left(-20\right)=-120$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the 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.

$$1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2$$

The solution is the pair that gives sum -7.

$$a=-15 b=8$$

Rewrite 6x^{2}-7x-20 as \left(6x^{2}-15x\right)+\left(8x-20\right).

$$\left(6x^{2}-15x\right)+\left(8x-20\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(2x-5\right)+4\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(3x+4\right)$$

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.

$$6x^{2}-7x-20=0$$

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.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 6\left(-20\right)}}{2\times 6}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 6\left(-20\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-7\right)±\sqrt{49-24\left(-20\right)}}{2\times 6}$$

Multiply -24 times -20.

$$x=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 6}$$

Add 49 to 480.

$$x=\frac{-\left(-7\right)±\sqrt{529}}{2\times 6}$$

Take the square root of 529.

$$x=\frac{-\left(-7\right)±23}{2\times 6}$$

The opposite of -7 is 7.

$$x=\frac{7±23}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{7±23}{12}$$

Now solve the equation x=\frac{7±23}{12} when ± is plus. Add 7 to 23.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{7±23}{12} when ± is minus. Subtract 23 from 7.

$$x=-\frac{16}{12}$$

Reduce the fraction \frac{-16}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -\frac{4}{3} for x_{2}.

$$6x^{2}-7x-20=6\left(x-\frac{5}{2}\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$6x^{2}-7x-20=6\left(x-\frac{5}{2}\right)\left(x+\frac{4}{3}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-7x-20=6\times \frac{2x-5}{2}\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-7x-20=6\times \frac{2x-5}{2}\times \frac{3x+4}{3}$$

Multiply \frac{2x-5}{2} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-7x-20=6\times \frac{\left(2x-5\right)\left(3x+4\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-7x-20=6\times \frac{\left(2x-5\right)\left(3x+4\right)}{6}$$

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

$$6x^{2}-7x-20=\left(2x-5\right)\left(3x+4\right)$$

Evaluate

$$\left(2x-5\right)\left(3x+4\right)$$