Solve 3x^2-19x+28

$$3 x ^ { 2 } - 19 x + 28$$

Factor

$$\left(x-4\right)\left(3x-7\right)$$

Solution Steps

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

$$a+b=-19 ab=3\times 28=84$$

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 84.

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

Calculate the sum for each pair.

$$-1-84=-85 -2-42=-44 -3-28=-31 -4-21=-25 -6-14=-20 -7-12=-19$$

The solution is the pair that gives sum -19.

$$a=-12 b=-7$$

Rewrite 3x^{2}-19x+28 as \left(3x^{2}-12x\right)+\left(-7x+28\right).

$$\left(3x^{2}-12x\right)+\left(-7x+28\right)$$

Factor out 3x in the first and -7 in the second group.

$$3x\left(x-4\right)-7\left(x-4\right)$$

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

$$\left(x-4\right)\left(3x-7\right)$$

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.

$$3x^{2}-19x+28=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 3\times 28}}{2\times 3}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 3\times 28}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-19\right)±\sqrt{361-12\times 28}}{2\times 3}$$

Multiply -12 times 28.

$$x=\frac{-\left(-19\right)±\sqrt{361-336}}{2\times 3}$$

Add 361 to -336.

$$x=\frac{-\left(-19\right)±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-\left(-19\right)±5}{2\times 3}$$

The opposite of -19 is 19.

$$x=\frac{19±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{19±5}{6}$$

Now solve the equation x=\frac{19±5}{6} when ± is plus. Add 19 to 5.

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

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{19±5}{6} when ± is minus. Subtract 5 from 19.

$$x=\frac{14}{6}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{3}$$

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{7}{3} for x_{2}.

$$3x^{2}-19x+28=3\left(x-4\right)\left(x-\frac{7}{3}\right)$$

Subtract \frac{7}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-19x+28=3\left(x-4\right)\times \frac{3x-7}{3}$$

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

$$3x^{2}-19x+28=\left(x-4\right)\left(3x-7\right)$$

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

$$x ^ 2 -\frac{19}{3}x +\frac{28}{3} = 0$$

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

$$r + s = \frac{19}{3} rs = \frac{28}{3}$$

Two numbers r and s sum up to \frac{19}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{19}{3} = \frac{19}{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>

$$r = \frac{19}{6} - u s = \frac{19}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{28}{3}

$$(\frac{19}{6} - u) (\frac{19}{6} + u) = \frac{28}{3}$$

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

$$\frac{361}{36} - u^2 = \frac{28}{3}$$

Simplify the expression by subtracting \frac{361}{36} on both sides

$$-u^2 = \frac{28}{3}-\frac{361}{36} = -\frac{25}{36}$$

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.

$$r =\frac{19}{6} - \frac{5}{6} = 2.333 s = \frac{19}{6} + \frac{5}{6} = 4.000$$

Evaluate

$$\left(x-4\right)\left(3x-7\right)$$