Solve b^2+14b+40

$$b ^ { 2 } + 14 b + 40$$

Factor

$$\left(b+4\right)\left(b+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb+40. To find p and q, set up a system to be solved.

$$p+q=14 pq=1\times 40=40$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 40.

$$1,40 2,20 4,10 5,8$$

Calculate the sum for each pair.

$$1+40=41 2+20=22 4+10=14 5+8=13$$

The solution is the pair that gives sum 14.

$$p=4 q=10$$

Rewrite b^{2}+14b+40 as \left(b^{2}+4b\right)+\left(10b+40\right).

$$\left(b^{2}+4b\right)+\left(10b+40\right)$$

Factor out b in the first and 10 in the second group.

$$b\left(b+4\right)+10\left(b+4\right)$$

Factor out common term b+4 by using distributive property.

$$\left(b+4\right)\left(b+10\right)$$

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.

$$b^{2}+14b+40=0$$

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.

$$b=\frac{-14±\sqrt{14^{2}-4\times 40}}{2}$$

Square 14.

$$b=\frac{-14±\sqrt{196-4\times 40}}{2}$$

Multiply -4 times 40.

$$b=\frac{-14±\sqrt{196-160}}{2}$$

Add 196 to -160.

$$b=\frac{-14±\sqrt{36}}{2}$$

Take the square root of 36.

$$b=\frac{-14±6}{2}$$

Now solve the equation b=\frac{-14±6}{2} when ± is plus. Add -14 to 6.

$$b=-\frac{8}{2}$$

Divide -8 by 2.

$$b=-4$$

Now solve the equation b=\frac{-14±6}{2} when ± is minus. Subtract 6 from -14.

$$b=-\frac{20}{2}$$

Divide -20 by 2.

$$b=-10$$

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 -10 for x_{2}.

$$b^{2}+14b+40=\left(b-\left(-4\right)\right)\left(b-\left(-10\right)\right)$$

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

$$b^{2}+14b+40=\left(b+4\right)\left(b+10\right)$$

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.

$$x ^ 2 +14x +40 = 0$$

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

$$r + s = -14 rs = 40$$

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 = 40

$$(-7 - u) (-7 + u) = 40$$

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

$$49 - u^2 = 40$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 40-49 = -9$$

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

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

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

$$r =-7 - 3 = -10 s = -7 + 3 = -4$$

Evaluate

$$\left(b+4\right)\left(b+10\right)$$