Solve 28k^2+13k-6

$$28 k ^ { 2 } + 13 k - 6$$

Factor

$$\left(7k-2\right)\left(4k+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 28k^{2}+ak+bk-6. To find a and b, set up a system to be solved.

$$a+b=13 ab=28\left(-6\right)=-168$$

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 -168.

$$-1,168 -2,84 -3,56 -4,42 -6,28 -7,24 -8,21 -12,14$$

Calculate the sum for each pair.

$$-1+168=167 -2+84=82 -3+56=53 -4+42=38 -6+28=22 -7+24=17 -8+21=13 -12+14=2$$

The solution is the pair that gives sum 13.

$$a=-8 b=21$$

Rewrite 28k^{2}+13k-6 as \left(28k^{2}-8k\right)+\left(21k-6\right).

$$\left(28k^{2}-8k\right)+\left(21k-6\right)$$

Factor out 4k in the first and 3 in the second group.

$$4k\left(7k-2\right)+3\left(7k-2\right)$$

Factor out common term 7k-2 by using distributive property.

$$\left(7k-2\right)\left(4k+3\right)$$

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.

$$28k^{2}+13k-6=0$$

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.

$$k=\frac{-13±\sqrt{13^{2}-4\times 28\left(-6\right)}}{2\times 28}$$

Square 13.

$$k=\frac{-13±\sqrt{169-4\times 28\left(-6\right)}}{2\times 28}$$

Multiply -4 times 28.

$$k=\frac{-13±\sqrt{169-112\left(-6\right)}}{2\times 28}$$

Multiply -112 times -6.

$$k=\frac{-13±\sqrt{169+672}}{2\times 28}$$

Add 169 to 672.

$$k=\frac{-13±\sqrt{841}}{2\times 28}$$

Take the square root of 841.

$$k=\frac{-13±29}{2\times 28}$$

Multiply 2 times 28.

$$k=\frac{-13±29}{56}$$

Now solve the equation k=\frac{-13±29}{56} when ± is plus. Add -13 to 29.

$$k=\frac{16}{56}$$

Reduce the fraction \frac{16}{56} to lowest terms by extracting and canceling out 8.

$$k=\frac{2}{7}$$

Now solve the equation k=\frac{-13±29}{56} when ± is minus. Subtract 29 from -13.

$$k=-\frac{42}{56}$$

Reduce the fraction \frac{-42}{56} to lowest terms by extracting and canceling out 14.

$$k=-\frac{3}{4}$$

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

$$28k^{2}+13k-6=28\left(k-\frac{2}{7}\right)\left(k-\left(-\frac{3}{4}\right)\right)$$

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

$$28k^{2}+13k-6=28\left(k-\frac{2}{7}\right)\left(k+\frac{3}{4}\right)$$

Subtract \frac{2}{7} from k by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$28k^{2}+13k-6=28\times \frac{7k-2}{7}\left(k+\frac{3}{4}\right)$$

Add \frac{3}{4} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$28k^{2}+13k-6=28\times \frac{7k-2}{7}\times \frac{4k+3}{4}$$

Multiply \frac{7k-2}{7} times \frac{4k+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$28k^{2}+13k-6=28\times \frac{\left(7k-2\right)\left(4k+3\right)}{7\times 4}$$

Multiply 7 times 4.

$$28k^{2}+13k-6=28\times \frac{\left(7k-2\right)\left(4k+3\right)}{28}$$

Cancel out 28, the greatest common factor in 28 and 28.

$$28k^{2}+13k-6=\left(7k-2\right)\left(4k+3\right)$$

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 28

$$x ^ 2 +\frac{13}{28}x -\frac{3}{14} = 0$$

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

$$r + s = -\frac{13}{28} rs = -\frac{3}{14}$$

Two numbers r and s sum up to -\frac{13}{28} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{28} = -\frac{13}{56}. 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>

$$r = -\frac{13}{56} - u s = -\frac{13}{56} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{14}

$$(-\frac{13}{56} - u) (-\frac{13}{56} + u) = -\frac{3}{14}$$

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

$$\frac{169}{3136} - u^2 = -\frac{3}{14}$$

Simplify the expression by subtracting \frac{169}{3136} on both sides

$$-u^2 = -\frac{3}{14}-\frac{169}{3136} = -\frac{841}{3136}$$

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

$$u^2 = \frac{841}{3136} u = \pm\sqrt{\frac{841}{3136}} = \pm \frac{29}{56}$$

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

$$r =-\frac{13}{56} - \frac{29}{56} = -0.750 s = -\frac{13}{56} + \frac{29}{56} = 0.286$$

Evaluate

$$\left(7k-2\right)\left(4k+3\right)$$