Solve 6x^2+1x-15

$$6 { x }^{ 2 } +1x-15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-15\right)=-90$$

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 -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 1.

$$a=-9 b=10$$

Rewrite 6x^{2}+x-15 as \left(6x^{2}-9x\right)+\left(10x-15\right).

$$\left(6x^{2}-9x\right)+\left(10x-15\right)$$

Factor out 3x in the first and 5 in the second group.

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

Factor out common term 2x-3 by using distributive property.

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

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.

$$6x^{2}+x-15=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 6\left(-15\right)}}{2\times 6}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 6\left(-15\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-1±\sqrt{1-24\left(-15\right)}}{2\times 6}$$

Multiply -24 times -15.

$$x=\frac{-1±\sqrt{1+360}}{2\times 6}$$

Add 1 to 360.

$$x=\frac{-1±\sqrt{361}}{2\times 6}$$

Take the square root of 361.

$$x=\frac{-1±19}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-1±19}{12}$$

Now solve the equation x=\frac{-1±19}{12} when ± is plus. Add -1 to 19.

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

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

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

Now solve the equation x=\frac{-1±19}{12} when ± is minus. Subtract 19 from -1.

$$x=-\frac{20}{12}$$

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{3}$$

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

$$6x^{2}+x-15=6\left(x-\frac{3}{2}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$6x^{2}+x-15=6\left(x-\frac{3}{2}\right)\left(x+\frac{5}{3}\right)$$

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

$$6x^{2}+x-15=6\times \frac{2x-3}{2}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}+x-15=6\times \frac{2x-3}{2}\times \frac{3x+5}{3}$$

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

$$6x^{2}+x-15=6\times \frac{\left(2x-3\right)\left(3x+5\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}+x-15=6\times \frac{\left(2x-3\right)\left(3x+5\right)}{6}$$

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

$$6x^{2}+x-15=\left(2x-3\right)\left(3x+5\right)$$

Evaluate

$$6x^{2}+x-15$$