Solve r^2+13r+36

$$r ^ { 2 } + 13 r + 36$$

Factor

$$\left(r+4\right)\left(r+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br+36. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\times 36=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=4 b=9$$

Rewrite r^{2}+13r+36 as \left(r^{2}+4r\right)+\left(9r+36\right).

$$\left(r^{2}+4r\right)+\left(9r+36\right)$$

Factor out r in the first and 9 in the second group.

$$r\left(r+4\right)+9\left(r+4\right)$$

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

$$\left(r+4\right)\left(r+9\right)$$

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.

$$r^{2}+13r+36=0$$

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.

$$r=\frac{-13±\sqrt{13^{2}-4\times 36}}{2}$$

Square 13.

$$r=\frac{-13±\sqrt{169-4\times 36}}{2}$$

Multiply -4 times 36.

$$r=\frac{-13±\sqrt{169-144}}{2}$$

Add 169 to -144.

$$r=\frac{-13±\sqrt{25}}{2}$$

Take the square root of 25.

$$r=\frac{-13±5}{2}$$

Now solve the equation r=\frac{-13±5}{2} when ± is plus. Add -13 to 5.

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

Divide -8 by 2.

$$r=-4$$

Now solve the equation r=\frac{-13±5}{2} when ± is minus. Subtract 5 from -13.

$$r=-\frac{18}{2}$$

Divide -18 by 2.

$$r=-9$$

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 -9 for x_{2}.

$$r^{2}+13r+36=\left(r-\left(-4\right)\right)\left(r-\left(-9\right)\right)$$

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

$$r^{2}+13r+36=\left(r+4\right)\left(r+9\right)$$

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.

$$x ^ 2 +13x +36 = 0$$

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

$$r + s = -13 rs = 36$$

Two numbers r and s sum up to -13 exactly when the average of the two numbers is \frac{1}{2}*-13 = -\frac{13}{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>

$$r = -\frac{13}{2} - u s = -\frac{13}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 36

$$(-\frac{13}{2} - u) (-\frac{13}{2} + u) = 36$$

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

$$\frac{169}{4} - u^2 = 36$$

Simplify the expression by subtracting \frac{169}{4} on both sides

$$-u^2 = 36-\frac{169}{4} = -\frac{25}{4}$$

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

$$u^2 = \frac{25}{4} u = \pm\sqrt{\frac{25}{4}} = \pm \frac{5}{2}$$

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

$$r =-\frac{13}{2} - \frac{5}{2} = -9 s = -\frac{13}{2} + \frac{5}{2} = -4$$

Evaluate

$$\left(r+4\right)\left(r+9\right)$$