Solve 3y^2+17y-28

$$3 { y }^{ 2 } +17y-28$$

Factor

$$\left(3y-4\right)\left(y+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by-28. To find a and b, set up a system to be solved.

$$a+b=17 ab=3\left(-28\right)=-84$$

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 -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 17.

$$a=-4 b=21$$

Rewrite 3y^{2}+17y-28 as \left(3y^{2}-4y\right)+\left(21y-28\right).

$$\left(3y^{2}-4y\right)+\left(21y-28\right)$$

Factor out y in the first and 7 in the second group.

$$y\left(3y-4\right)+7\left(3y-4\right)$$

Factor out common term 3y-4 by using distributive property.

$$\left(3y-4\right)\left(y+7\right)$$

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.

$$3y^{2}+17y-28=0$$

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.

$$y=\frac{-17±\sqrt{17^{2}-4\times 3\left(-28\right)}}{2\times 3}$$

Square 17.

$$y=\frac{-17±\sqrt{289-4\times 3\left(-28\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-17±\sqrt{289-12\left(-28\right)}}{2\times 3}$$

Multiply -12 times -28.

$$y=\frac{-17±\sqrt{289+336}}{2\times 3}$$

Add 289 to 336.

$$y=\frac{-17±\sqrt{625}}{2\times 3}$$

Take the square root of 625.

$$y=\frac{-17±25}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-17±25}{6}$$

Now solve the equation y=\frac{-17±25}{6} when ± is plus. Add -17 to 25.

$$y=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{4}{3}$$

Now solve the equation y=\frac{-17±25}{6} when ± is minus. Subtract 25 from -17.

$$y=-\frac{42}{6}$$

Divide -42 by 6.

$$y=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -7 for x_{2}.

$$3y^{2}+17y-28=3\left(y-\frac{4}{3}\right)\left(y-\left(-7\right)\right)$$

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

$$3y^{2}+17y-28=3\left(y-\frac{4}{3}\right)\left(y+7\right)$$

Subtract \frac{4}{3} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}+17y-28=3\times \frac{3y-4}{3}\left(y+7\right)$$

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

$$3y^{2}+17y-28=\left(3y-4\right)\left(y+7\right)$$

Evaluate

$$\left(3y-4\right)\left(y+7\right)$$