Solve 6y+9y^2-15

$$6 y + 9 y ^ { 2 } - 15$$

Factor

$$3\left(y-1\right)\left(3y+5\right)$$

Solution Steps

Factor out 3.

$$3\left(2y+3y^{2}-5\right)$$

Consider 2y+3y^{2}-5. Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3y^{2}+2y-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-5\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 3y^{2}+2y-5 as \left(3y^{2}-3y\right)+\left(5y-5\right).

$$\left(3y^{2}-3y\right)+\left(5y-5\right)$$

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

$$3y\left(y-1\right)+5\left(y-1\right)$$

Factor out common term y-1 by using distributive property.

$$\left(y-1\right)\left(3y+5\right)$$

Rewrite the complete factored expression.

$$3\left(y-1\right)\left(3y+5\right)$$

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.

$$9y^{2}+6y-15=0$$

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.

$$y=\frac{-6±\sqrt{6^{2}-4\times 9\left(-15\right)}}{2\times 9}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\times 9\left(-15\right)}}{2\times 9}$$

Multiply -4 times 9.

$$y=\frac{-6±\sqrt{36-36\left(-15\right)}}{2\times 9}$$

Multiply -36 times -15.

$$y=\frac{-6±\sqrt{36+540}}{2\times 9}$$

Add 36 to 540.

$$y=\frac{-6±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$y=\frac{-6±24}{2\times 9}$$

Multiply 2 times 9.

$$y=\frac{-6±24}{18}$$

Now solve the equation y=\frac{-6±24}{18} when ± is plus. Add -6 to 24.

$$y=\frac{18}{18}$$

Divide 18 by 18.

$$y=1$$

Now solve the equation y=\frac{-6±24}{18} when ± is minus. Subtract 24 from -6.

$$y=-\frac{30}{18}$$

Reduce the fraction \frac{-30}{18} to lowest terms by extracting and canceling out 6.

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

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

$$9y^{2}+6y-15=9\left(y-1\right)\left(y-\left(-\frac{5}{3}\right)\right)$$

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

$$9y^{2}+6y-15=9\left(y-1\right)\left(y+\frac{5}{3}\right)$$

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

$$9y^{2}+6y-15=9\left(y-1\right)\times \frac{3y+5}{3}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$9y^{2}+6y-15=3\left(y-1\right)\left(3y+5\right)$$

Evaluate

$$3\left(y-1\right)\left(3y+5\right)$$