Solve 3x^2-20x-63

$$3 { x }^{ 2 } -20x-63$$

Factor

$$\left(x-9\right)\left(3x+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-63. To find a and b, set up a system to be solved.

$$a+b=-20 ab=3\left(-63\right)=-189$$

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 -189.

$$1,-189 3,-63 7,-27 9,-21$$

Calculate the sum for each pair.

$$1-189=-188 3-63=-60 7-27=-20 9-21=-12$$

The solution is the pair that gives sum -20.

$$a=-27 b=7$$

Rewrite 3x^{2}-20x-63 as \left(3x^{2}-27x\right)+\left(7x-63\right).

$$\left(3x^{2}-27x\right)+\left(7x-63\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(x-9\right)+7\left(x-9\right)$$

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

$$\left(x-9\right)\left(3x+7\right)$$

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.

$$3x^{2}-20x-63=0$$

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.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 3\left(-63\right)}}{2\times 3}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 3\left(-63\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-20\right)±\sqrt{400-12\left(-63\right)}}{2\times 3}$$

Multiply -12 times -63.

$$x=\frac{-\left(-20\right)±\sqrt{400+756}}{2\times 3}$$

Add 400 to 756.

$$x=\frac{-\left(-20\right)±\sqrt{1156}}{2\times 3}$$

Take the square root of 1156.

$$x=\frac{-\left(-20\right)±34}{2\times 3}$$

The opposite of -20 is 20.

$$x=\frac{20±34}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{20±34}{6}$$

Now solve the equation x=\frac{20±34}{6} when ± is plus. Add 20 to 34.

$$x=\frac{54}{6}$$

Divide 54 by 6.

$$x=9$$

Now solve the equation x=\frac{20±34}{6} when ± is minus. Subtract 34 from 20.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and -\frac{7}{3} for x_{2}.

$$3x^{2}-20x-63=3\left(x-9\right)\left(x-\left(-\frac{7}{3}\right)\right)$$

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

$$3x^{2}-20x-63=3\left(x-9\right)\left(x+\frac{7}{3}\right)$$

Add \frac{7}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-20x-63=3\left(x-9\right)\times \frac{3x+7}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-20x-63=\left(x-9\right)\left(3x+7\right)$$

Evaluate

$$\left(x-9\right)\left(3x+7\right)$$