Solve 10x^2-3x-27=

$$10 { x }^{ 2 } -3x-27=$$

Factor

$$\left(5x-9\right)\left(2x+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-3 ab=10\left(-27\right)=-270$$

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 -270.

$$1,-270 2,-135 3,-90 5,-54 6,-45 9,-30 10,-27 15,-18$$

Calculate the sum for each pair.

$$1-270=-269 2-135=-133 3-90=-87 5-54=-49 6-45=-39 9-30=-21 10-27=-17 15-18=-3$$

The solution is the pair that gives sum -3.

$$a=-18 b=15$$

Rewrite 10x^{2}-3x-27 as \left(10x^{2}-18x\right)+\left(15x-27\right).

$$\left(10x^{2}-18x\right)+\left(15x-27\right)$$

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

$$2x\left(5x-9\right)+3\left(5x-9\right)$$

Factor out common term 5x-9 by using distributive property.

$$\left(5x-9\right)\left(2x+3\right)$$

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.

$$10x^{2}-3x-27=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 10\left(-27\right)}}{2\times 10}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 10\left(-27\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-3\right)±\sqrt{9-40\left(-27\right)}}{2\times 10}$$

Multiply -40 times -27.

$$x=\frac{-\left(-3\right)±\sqrt{9+1080}}{2\times 10}$$

Add 9 to 1080.

$$x=\frac{-\left(-3\right)±\sqrt{1089}}{2\times 10}$$

Take the square root of 1089.

$$x=\frac{-\left(-3\right)±33}{2\times 10}$$

The opposite of -3 is 3.

$$x=\frac{3±33}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{3±33}{20}$$

Now solve the equation x=\frac{3±33}{20} when ± is plus. Add 3 to 33.

$$x=\frac{36}{20}$$

Reduce the fraction \frac{36}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{9}{5}$$

Now solve the equation x=\frac{3±33}{20} when ± is minus. Subtract 33 from 3.

$$x=-\frac{30}{20}$$

Reduce the fraction \frac{-30}{20} to lowest terms by extracting and canceling out 10.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{5} for x_{1} and -\frac{3}{2} for x_{2}.

$$10x^{2}-3x-27=10\left(x-\frac{9}{5}\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$10x^{2}-3x-27=10\left(x-\frac{9}{5}\right)\left(x+\frac{3}{2}\right)$$

Subtract \frac{9}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-3x-27=10\times \frac{5x-9}{5}\left(x+\frac{3}{2}\right)$$

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

$$10x^{2}-3x-27=10\times \frac{5x-9}{5}\times \frac{2x+3}{2}$$

Multiply \frac{5x-9}{5} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}-3x-27=10\times \frac{\left(5x-9\right)\left(2x+3\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}-3x-27=10\times \frac{\left(5x-9\right)\left(2x+3\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}-3x-27=\left(5x-9\right)\left(2x+3\right)$$

Evaluate

$$\left(5x-9\right)\left(2x+3\right)$$