Solve t^2-6t+8

$$t ^ { 2 } - 6 t + 8$$

Factor

$$\left(t-4\right)\left(t-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite t^{2}-6t+8 as \left(t^{2}-4t\right)+\left(-2t+8\right).

$$\left(t^{2}-4t\right)+\left(-2t+8\right)$$

Factor out t in the first and -2 in the second group.

$$t\left(t-4\right)-2\left(t-4\right)$$

Factor out common term t-4 by using distributive property.

$$\left(t-4\right)\left(t-2\right)$$

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.

$$t^{2}-6t+8=0$$

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.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8}}{2}$$

Square -6.

$$t=\frac{-\left(-6\right)±\sqrt{36-4\times 8}}{2}$$

Multiply -4 times 8.

$$t=\frac{-\left(-6\right)±\sqrt{36-32}}{2}$$

Add 36 to -32.

$$t=\frac{-\left(-6\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$t=\frac{-\left(-6\right)±2}{2}$$

The opposite of -6 is 6.

$$t=\frac{6±2}{2}$$

Now solve the equation t=\frac{6±2}{2} when ± is plus. Add 6 to 2.

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

Divide 8 by 2.

$$t=4$$

Now solve the equation t=\frac{6±2}{2} when ± is minus. Subtract 2 from 6.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

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

$$t^{2}-6t+8=\left(t-4\right)\left(t-2\right)$$

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.

$$x ^ 2 -6x +8 = 0$$

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

$$r + s = 6 rs = 8$$

Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 3. 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>

$$r = 3 - u s = 3 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 8

$$(3 - u) (3 + u) = 8$$

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

$$9 - u^2 = 8$$

Simplify the expression by subtracting 9 on both sides

$$-u^2 = 8-9 = -1$$

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.

$$r =3 - 1 = 2 s = 3 + 1 = 4$$

Evaluate

$$\left(t-4\right)\left(t-2\right)$$