Solve 6w^2-28w+16

$$6 w ^ { 2 } - 28 w + 16$$

Factor

$$2\left(w-4\right)\left(3w-2\right)$$

Solution Steps

Factor out 2.

$$2\left(3w^{2}-14w+8\right)$$

Consider 3w^{2}-14w+8. Factor the expression by grouping. First, the expression needs to be rewritten as 3w^{2}+aw+bw+8. To find a and b, set up a system to be solved.

$$a+b=-14 ab=3\times 8=24$$

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 -14.

$$a=-12 b=-2$$

Rewrite 3w^{2}-14w+8 as \left(3w^{2}-12w\right)+\left(-2w+8\right).

$$\left(3w^{2}-12w\right)+\left(-2w+8\right)$$

Factor out 3w in the first and -2 in the second group.

$$3w\left(w-4\right)-2\left(w-4\right)$$

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

$$\left(w-4\right)\left(3w-2\right)$$

Rewrite the complete factored expression.

$$2\left(w-4\right)\left(3w-2\right)$$

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.

$$6w^{2}-28w+16=0$$

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.

$$w=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 6\times 16}}{2\times 6}$$

Square -28.

$$w=\frac{-\left(-28\right)±\sqrt{784-4\times 6\times 16}}{2\times 6}$$

Multiply -4 times 6.

$$w=\frac{-\left(-28\right)±\sqrt{784-24\times 16}}{2\times 6}$$

Multiply -24 times 16.

$$w=\frac{-\left(-28\right)±\sqrt{784-384}}{2\times 6}$$

Add 784 to -384.

$$w=\frac{-\left(-28\right)±\sqrt{400}}{2\times 6}$$

Take the square root of 400.

$$w=\frac{-\left(-28\right)±20}{2\times 6}$$

The opposite of -28 is 28.

$$w=\frac{28±20}{2\times 6}$$

Multiply 2 times 6.

$$w=\frac{28±20}{12}$$

Now solve the equation w=\frac{28±20}{12} when ± is plus. Add 28 to 20.

$$w=\frac{48}{12}$$

Divide 48 by 12.

$$w=4$$

Now solve the equation w=\frac{28±20}{12} when ± is minus. Subtract 20 from 28.

$$w=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

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 \frac{2}{3} for x_{2}.

$$6w^{2}-28w+16=6\left(w-4\right)\left(w-\frac{2}{3}\right)$$

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

$$6w^{2}-28w+16=6\left(w-4\right)\times \frac{3w-2}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$6w^{2}-28w+16=2\left(w-4\right)\left(3w-2\right)$$

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 6

$$x ^ 2 -\frac{14}{3}x +\frac{8}{3} = 0$$

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

$$r + s = \frac{14}{3} rs = \frac{8}{3}$$

Two numbers r and s sum up to \frac{14}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{14}{3} = \frac{7}{3}. 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>

$$r = \frac{7}{3} - u s = \frac{7}{3} + u$$

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

$$(\frac{7}{3} - u) (\frac{7}{3} + u) = \frac{8}{3}$$

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

$$\frac{49}{9} - u^2 = \frac{8}{3}$$

Simplify the expression by subtracting \frac{49}{9} on both sides

$$-u^2 = \frac{8}{3}-\frac{49}{9} = -\frac{25}{9}$$

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

$$u^2 = \frac{25}{9} u = \pm\sqrt{\frac{25}{9}} = \pm \frac{5}{3}$$

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

$$r =\frac{7}{3} - \frac{5}{3} = 0.667 s = \frac{7}{3} + \frac{5}{3} = 4$$

Evaluate

$$2\left(w-4\right)\left(3w-2\right)$$