Solve b^2-8b+15

$$b ^ { 2 } - 8 b + 15$$

Factor

$$\left(b-5\right)\left(b-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb+15. To find p and q, set up a system to be solved.

$$p+q=-8 pq=1\times 15=15$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$p=-5 q=-3$$

Rewrite b^{2}-8b+15 as \left(b^{2}-5b\right)+\left(-3b+15\right).

$$\left(b^{2}-5b\right)+\left(-3b+15\right)$$

Factor out b in the first and -3 in the second group.

$$b\left(b-5\right)-3\left(b-5\right)$$

Factor out common term b-5 by using distributive property.

$$\left(b-5\right)\left(b-3\right)$$

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.

$$b^{2}-8b+15=0$$

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.

$$b=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Square -8.

$$b=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Multiply -4 times 15.

$$b=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Add 64 to -60.

$$b=\frac{-\left(-8\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$b=\frac{-\left(-8\right)±2}{2}$$

The opposite of -8 is 8.

$$b=\frac{8±2}{2}$$

Now solve the equation b=\frac{8±2}{2} when ± is plus. Add 8 to 2.

$$b=\frac{10}{2}$$

Divide 10 by 2.

$$b=5$$

Now solve the equation b=\frac{8±2}{2} when ± is minus. Subtract 2 from 8.

$$b=\frac{6}{2}$$

Divide 6 by 2.

$$b=3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and 3 for x_{2}.

$$b^{2}-8b+15=\left(b-5\right)\left(b-3\right)$$

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.

$$x ^ 2 -8x +15 = 0$$

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

$$r + s = 8 rs = 15$$

Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 4. 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>

$$r = 4 - u s = 4 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 15

$$(4 - u) (4 + u) = 15$$

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

$$16 - u^2 = 15$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = 15-16 = -1$$

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.

$$r =4 - 1 = 3 s = 4 + 1 = 5$$

Evaluate

$$\left(b-5\right)\left(b-3\right)$$