Solve t^2+2t-48

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

Factor

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

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.

$$a+b=2 ab=1\left(-48\right)=-48$$

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 -48.

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

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 2.

$$a=-6 b=8$$

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

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

Factor out t in the first and 8 in the second group.

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

Factor out common term t-6 by using distributive property.

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

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.

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

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.

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

Square 2.

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

Multiply -4 times -48.

$$t=\frac{-2±\sqrt{4+192}}{2}$$

Add 4 to 192.

$$t=\frac{-2±\sqrt{196}}{2}$$

Take the square root of 196.

$$t=\frac{-2±14}{2}$$

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

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

Divide 12 by 2.

$$t=6$$

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

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

Divide -16 by 2.

$$t=-8$$

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

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

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

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

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.

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

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

$$r + s = -2 rs = -48$$

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

$$r = -1 - u s = -1 + u$$

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

$$(-1 - u) (-1 + u) = -48$$

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

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

Simplify the expression by subtracting 1 on both sides

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

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

$$u^2 = 49 u = \pm\sqrt{49} = \pm 7$$

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

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

Evaluate

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