Solve q^2-11q+28

$$q ^ { 2 } - 11 q + 28$$

Factor

$$\left(q-7\right)\left(q-4\right)$$

Solution Steps

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

$$a+b=-11 ab=1\times 28=28$$

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 28.

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

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -11.

$$a=-7 b=-4$$

Rewrite q^{2}-11q+28 as \left(q^{2}-7q\right)+\left(-4q+28\right).

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

Factor out q in the first and -4 in the second group.

$$q\left(q-7\right)-4\left(q-7\right)$$

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

$$\left(q-7\right)\left(q-4\right)$$

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.

$$q^{2}-11q+28=0$$

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.

$$q=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}$$

Square -11.

$$q=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}$$

Multiply -4 times 28.

$$q=\frac{-\left(-11\right)±\sqrt{121-112}}{2}$$

Add 121 to -112.

$$q=\frac{-\left(-11\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$q=\frac{-\left(-11\right)±3}{2}$$

The opposite of -11 is 11.

$$q=\frac{11±3}{2}$$

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

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

Divide 14 by 2.

$$q=7$$

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

$$q=\frac{8}{2}$$

Divide 8 by 2.

$$q=4$$

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}.

$$q^{2}-11q+28=\left(q-7\right)\left(q-4\right)$$

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.

$$x ^ 2 -11x +28 = 0$$

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

$$r + s = 11 rs = 28$$

Two numbers r and s sum up to 11 exactly when the average of the two numbers is \frac{1}{2}*11 = \frac{11}{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>

$$r = \frac{11}{2} - u s = \frac{11}{2} + u$$

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

$$(\frac{11}{2} - u) (\frac{11}{2} + u) = 28$$

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

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

Simplify the expression by subtracting \frac{121}{4} on both sides

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

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

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

Evaluate

$$\left(q-7\right)\left(q-4\right)$$