Solve p^2+19p+18

$$p ^ { 2 } + 19 p + 18$$

Factor

$$\left(p+1\right)\left(p+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+18. To find a and b, set up a system to be solved.

$$a+b=19 ab=1\times 18=18$$

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 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 19.

$$a=1 b=18$$

Rewrite p^{2}+19p+18 as \left(p^{2}+p\right)+\left(18p+18\right).

$$\left(p^{2}+p\right)+\left(18p+18\right)$$

Factor out p in the first and 18 in the second group.

$$p\left(p+1\right)+18\left(p+1\right)$$

Factor out common term p+1 by using distributive property.

$$\left(p+1\right)\left(p+18\right)$$

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.

$$p^{2}+19p+18=0$$

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.

$$p=\frac{-19±\sqrt{19^{2}-4\times 18}}{2}$$

Square 19.

$$p=\frac{-19±\sqrt{361-4\times 18}}{2}$$

Multiply -4 times 18.

$$p=\frac{-19±\sqrt{361-72}}{2}$$

Add 361 to -72.

$$p=\frac{-19±\sqrt{289}}{2}$$

Take the square root of 289.

$$p=\frac{-19±17}{2}$$

Now solve the equation p=\frac{-19±17}{2} when ± is plus. Add -19 to 17.

$$p=-\frac{2}{2}$$

Divide -2 by 2.

$$p=-1$$

Now solve the equation p=\frac{-19±17}{2} when ± is minus. Subtract 17 from -19.

$$p=-\frac{36}{2}$$

Divide -36 by 2.

$$p=-18$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -18 for x_{2}.

$$p^{2}+19p+18=\left(p-\left(-1\right)\right)\left(p-\left(-18\right)\right)$$

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

$$p^{2}+19p+18=\left(p+1\right)\left(p+18\right)$$

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.

$$x ^ 2 +19x +18 = 0$$

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

$$r + s = -19 rs = 18$$

Two numbers r and s sum up to -19 exactly when the average of the two numbers is \frac{1}{2}*-19 = -\frac{19}{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>

$$r = -\frac{19}{2} - u s = -\frac{19}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 18

$$(-\frac{19}{2} - u) (-\frac{19}{2} + u) = 18$$

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

$$\frac{361}{4} - u^2 = 18$$

Simplify the expression by subtracting \frac{361}{4} on both sides

$$-u^2 = 18-\frac{361}{4} = -\frac{289}{4}$$

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

$$u^2 = \frac{289}{4} u = \pm\sqrt{\frac{289}{4}} = \pm \frac{17}{2}$$

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

$$r =-\frac{19}{2} - \frac{17}{2} = -18 s = -\frac{19}{2} + \frac{17}{2} = -1$$

Evaluate

$$\left(p+1\right)\left(p+18\right)$$