Solve 3y^2-11y+8

$$3 y ^ { 2 } - 11 y + 8$$

Factor

$$\left(y-1\right)\left(3y-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 8=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite 3y^{2}-11y+8 as \left(3y^{2}-8y\right)+\left(-3y+8\right).

$$\left(3y^{2}-8y\right)+\left(-3y+8\right)$$

Factor out y in the first and -1 in the second group.

$$y\left(3y-8\right)-\left(3y-8\right)$$

Factor out common term 3y-8 by using distributive property.

$$\left(3y-8\right)\left(y-1\right)$$

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.

$$3y^{2}-11y+8=0$$

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.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 8}}{2\times 3}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-\left(-11\right)±\sqrt{121-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$y=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 3}$$

Add 121 to -96.

$$y=\frac{-\left(-11\right)±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$y=\frac{-\left(-11\right)±5}{2\times 3}$$

The opposite of -11 is 11.

$$y=\frac{11±5}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{11±5}{6}$$

Now solve the equation y=\frac{11±5}{6} when ± is plus. Add 11 to 5.

$$y=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$y=\frac{8}{3}$$

Now solve the equation y=\frac{11±5}{6} when ± is minus. Subtract 5 from 11.

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

Divide 6 by 6.

$$y=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and 1 for x_{2}.

$$3y^{2}-11y+8=3\left(y-\frac{8}{3}\right)\left(y-1\right)$$

Subtract \frac{8}{3} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3y^{2}-11y+8=3\times \frac{3y-8}{3}\left(y-1\right)$$

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

$$3y^{2}-11y+8=\left(3y-8\right)\left(y-1\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3

$$x ^ 2 -\frac{11}{3}x +\frac{8}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{11}{3} rs = \frac{8}{3}$$

Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{6}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{11}{6} - u s = \frac{11}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{3}

$$(\frac{11}{6} - u) (\frac{11}{6} + u) = \frac{8}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{121}{36} - u^2 = \frac{8}{3}$$

Simplify the expression by subtracting \frac{121}{36} on both sides

$$-u^2 = \frac{8}{3}-\frac{121}{36} = -\frac{25}{36}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{11}{6} - \frac{5}{6} = 1 s = \frac{11}{6} + \frac{5}{6} = 2.667$$

Evaluate

$$\left(y-1\right)\left(3y-8\right)$$