Solve 6x^2-27x-54

$$6 { x }^{ 2 } -27x-54$$

Factor

$$3\left(x-6\right)\left(2x+3\right)$$

Solution Steps

Factor out 3.

$$3\left(2x^{2}-9x-18\right)$$

Consider 2x^{2}-9x-18. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=2\left(-18\right)=-36$$

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 -36.

$$1,-36 2,-18 3,-12 4,-9 6,-6$$

Calculate the sum for each pair.

$$1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0$$

The solution is the pair that gives sum -9.

$$a=-12 b=3$$

Rewrite 2x^{2}-9x-18 as \left(2x^{2}-12x\right)+\left(3x-18\right).

$$\left(2x^{2}-12x\right)+\left(3x-18\right)$$

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

$$2x\left(x-6\right)+3\left(x-6\right)$$

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

$$\left(x-6\right)\left(2x+3\right)$$

Rewrite the complete factored expression.

$$3\left(x-6\right)\left(2x+3\right)$$

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.

$$6x^{2}-27x-54=0$$

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.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 6\left(-54\right)}}{2\times 6}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 6\left(-54\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-27\right)±\sqrt{729-24\left(-54\right)}}{2\times 6}$$

Multiply -24 times -54.

$$x=\frac{-\left(-27\right)±\sqrt{729+1296}}{2\times 6}$$

Add 729 to 1296.

$$x=\frac{-\left(-27\right)±\sqrt{2025}}{2\times 6}$$

Take the square root of 2025.

$$x=\frac{-\left(-27\right)±45}{2\times 6}$$

The opposite of -27 is 27.

$$x=\frac{27±45}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{27±45}{12}$$

Now solve the equation x=\frac{27±45}{12} when ± is plus. Add 27 to 45.

$$x=\frac{72}{12}$$

Divide 72 by 12.

$$x=6$$

Now solve the equation x=\frac{27±45}{12} when ± is minus. Subtract 45 from 27.

$$x=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and -\frac{3}{2} for x_{2}.

$$6x^{2}-27x-54=6\left(x-6\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$6x^{2}-27x-54=6\left(x-6\right)\left(x+\frac{3}{2}\right)$$

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

$$6x^{2}-27x-54=6\left(x-6\right)\times \frac{2x+3}{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$6x^{2}-27x-54=3\left(x-6\right)\left(2x+3\right)$$

Evaluate

$$3\left(x-6\right)\left(2x+3\right)$$