Solve t^2-14t+48

$$t ^ { 2 } - 14 t + 48$$

Factor

$$\left(t-8\right)\left(t-6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+48. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 48=48$$

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 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -14.

$$a=-8 b=-6$$

Rewrite t^{2}-14t+48 as \left(t^{2}-8t\right)+\left(-6t+48\right).

$$\left(t^{2}-8t\right)+\left(-6t+48\right)$$

Factor out t in the first and -6 in the second group.

$$t\left(t-8\right)-6\left(t-8\right)$$

Factor out common term t-8 by using distributive property.

$$\left(t-8\right)\left(t-6\right)$$

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.

$$t^{2}-14t+48=0$$

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.

$$t=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 48}}{2}$$

Square -14.

$$t=\frac{-\left(-14\right)±\sqrt{196-4\times 48}}{2}$$

Multiply -4 times 48.

$$t=\frac{-\left(-14\right)±\sqrt{196-192}}{2}$$

Add 196 to -192.

$$t=\frac{-\left(-14\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$t=\frac{-\left(-14\right)±2}{2}$$

The opposite of -14 is 14.

$$t=\frac{14±2}{2}$$

Now solve the equation t=\frac{14±2}{2} when ± is plus. Add 14 to 2.

$$t=\frac{16}{2}$$

Divide 16 by 2.

$$t=8$$

Now solve the equation t=\frac{14±2}{2} when ± is minus. Subtract 2 from 14.

$$t=\frac{12}{2}$$

Divide 12 by 2.

$$t=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 6 for x_{2}.

$$t^{2}-14t+48=\left(t-8\right)\left(t-6\right)$$

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.

$$x ^ 2 -14x +48 = 0$$

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

$$r + s = 14 rs = 48$$

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

$$r = 7 - u s = 7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 48

$$(7 - u) (7 + u) = 48$$

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

$$49 - u^2 = 48$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 48-49 = -1$$

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

$$u^2 = 1 u = \pm\sqrt{1} = \pm 1$$

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

$$r =7 - 1 = 6 s = 7 + 1 = 8$$

Evaluate

$$\left(t-8\right)\left(t-6\right)$$