Solve b^2+b-20=

$$b ^ { 2 } + b - 20 =$$

Factor

$$\left(b-4\right)\left(b+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb-20. To find p and q, set up a system to be solved.

$$p+q=1 pq=1\left(-20\right)=-20$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$p=-4 q=5$$

Rewrite b^{2}+b-20 as \left(b^{2}-4b\right)+\left(5b-20\right).

$$\left(b^{2}-4b\right)+\left(5b-20\right)$$

Factor out b in the first and 5 in the second group.

$$b\left(b-4\right)+5\left(b-4\right)$$

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

$$\left(b-4\right)\left(b+5\right)$$

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.

$$b^{2}+b-20=0$$

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.

$$b=\frac{-1±\sqrt{1^{2}-4\left(-20\right)}}{2}$$

Square 1.

$$b=\frac{-1±\sqrt{1-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$b=\frac{-1±\sqrt{1+80}}{2}$$

Add 1 to 80.

$$b=\frac{-1±\sqrt{81}}{2}$$

Take the square root of 81.

$$b=\frac{-1±9}{2}$$

Now solve the equation b=\frac{-1±9}{2} when ± is plus. Add -1 to 9.

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

Divide 8 by 2.

$$b=4$$

Now solve the equation b=\frac{-1±9}{2} when ± is minus. Subtract 9 from -1.

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

Divide -10 by 2.

$$b=-5$$

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 -5 for x_{2}.

$$b^{2}+b-20=\left(b-4\right)\left(b-\left(-5\right)\right)$$

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

$$b^{2}+b-20=\left(b-4\right)\left(b+5\right)$$

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.

$$x ^ 2 +1x -20 = 0$$

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

$$r + s = -1 rs = -20$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -20

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -20$$

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

$$\frac{1}{4} - u^2 = -20$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -20-\frac{1}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =-\frac{1}{2} - \frac{9}{2} = -5 s = -\frac{1}{2} + \frac{9}{2} = 4$$

Evaluate

$$\left(b-4\right)\left(b+5\right)$$