Solve k^2-3k-28

$$k ^ { 2 } - 3 k - 28$$

Factor

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

Solution Steps

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

$$a+b=-3 ab=1\left(-28\right)=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -28.

$$1,-28 2,-14 4,-7$$

Calculate the sum for each pair.

$$1-28=-27 2-14=-12 4-7=-3$$

The solution is the pair that gives sum -3.

$$a=-7 b=4$$

Rewrite k^{2}-3k-28 as \left(k^{2}-7k\right)+\left(4k-28\right).

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

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

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

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

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

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.

$$k^{2}-3k-28=0$$

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.

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

Square -3.

$$k=\frac{-\left(-3\right)±\sqrt{9-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$k=\frac{-\left(-3\right)±\sqrt{9+112}}{2}$$

Add 9 to 112.

$$k=\frac{-\left(-3\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$k=\frac{-\left(-3\right)±11}{2}$$

The opposite of -3 is 3.

$$k=\frac{3±11}{2}$$

Now solve the equation k=\frac{3±11}{2} when ± is plus. Add 3 to 11.

$$k=\frac{14}{2}$$

Divide 14 by 2.

$$k=7$$

Now solve the equation k=\frac{3±11}{2} when ± is minus. Subtract 11 from 3.

$$k=-\frac{8}{2}$$

Divide -8 by 2.

$$k=-4$$

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

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

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

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

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.

$$x ^ 2 -3x -28 = 0$$

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

$$r + s = 3 rs = -28$$

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

$$r = \frac{3}{2} - u s = \frac{3}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -28

$$(\frac{3}{2} - u) (\frac{3}{2} + u) = -28$$

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

$$\frac{9}{4} - u^2 = -28$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -28-\frac{9}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =\frac{3}{2} - \frac{11}{2} = -4 s = \frac{3}{2} + \frac{11}{2} = 7$$

Evaluate

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