Solve 3y^2-21y+36

$$3 { y }^{ 2 } -21y+36$$

Factor

$$3\left(y-4\right)\left(y-3\right)$$

Solution Steps

Factor out 3.

$$3\left(y^{2}-7y+12\right)$$

Consider y^{2}-7y+12. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+12. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -7.

$$a=-4 b=-3$$

Rewrite y^{2}-7y+12 as \left(y^{2}-4y\right)+\left(-3y+12\right).

$$\left(y^{2}-4y\right)+\left(-3y+12\right)$$

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

$$y\left(y-4\right)-3\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y-3\right)$$

Rewrite the complete factored expression.

$$3\left(y-4\right)\left(y-3\right)$$

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.

$$3y^{2}-21y+36=0$$

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.

$$y=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 3\times 36}}{2\times 3}$$

Square -21.

$$y=\frac{-\left(-21\right)±\sqrt{441-4\times 3\times 36}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-21\right)±\sqrt{441-12\times 36}}{2\times 3}$$

Multiply -12 times 36.

$$y=\frac{-\left(-21\right)±\sqrt{441-432}}{2\times 3}$$

Add 441 to -432.

$$y=\frac{-\left(-21\right)±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$y=\frac{-\left(-21\right)±3}{2\times 3}$$

The opposite of -21 is 21.

$$y=\frac{21±3}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{21±3}{6}$$

Now solve the equation y=\frac{21±3}{6} when ± is plus. Add 21 to 3.

$$y=\frac{24}{6}$$

Divide 24 by 6.

$$y=4$$

Now solve the equation y=\frac{21±3}{6} when ± is minus. Subtract 3 from 21.

$$y=\frac{18}{6}$$

Divide 18 by 6.

$$y=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and 3 for x_{2}.

$$3y^{2}-21y+36=3\left(y-4\right)\left(y-3\right)$$

Evaluate

$$3\left(y-4\right)\left(y-3\right)$$