Solve h^2+9h+18

$$h ^ { 2 } + 9 h + 18$$

Factor

$$\left(h+3\right)\left(h+6\right)$$

Solution Steps

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

$$a+b=9 ab=1\times 18=18$$

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 9.

$$a=3 b=6$$

Rewrite h^{2}+9h+18 as \left(h^{2}+3h\right)+\left(6h+18\right).

$$\left(h^{2}+3h\right)+\left(6h+18\right)$$

Factor out h in the first and 6 in the second group.

$$h\left(h+3\right)+6\left(h+3\right)$$

Factor out common term h+3 by using distributive property.

$$\left(h+3\right)\left(h+6\right)$$

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.

$$h^{2}+9h+18=0$$

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.

$$h=\frac{-9±\sqrt{9^{2}-4\times 18}}{2}$$

Square 9.

$$h=\frac{-9±\sqrt{81-4\times 18}}{2}$$

Multiply -4 times 18.

$$h=\frac{-9±\sqrt{81-72}}{2}$$

Add 81 to -72.

$$h=\frac{-9±\sqrt{9}}{2}$$

Take the square root of 9.

$$h=\frac{-9±3}{2}$$

Now solve the equation h=\frac{-9±3}{2} when ± is plus. Add -9 to 3.

$$h=-\frac{6}{2}$$

Divide -6 by 2.

$$h=-3$$

Now solve the equation h=\frac{-9±3}{2} when ± is minus. Subtract 3 from -9.

$$h=-\frac{12}{2}$$

Divide -12 by 2.

$$h=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and -6 for x_{2}.

$$h^{2}+9h+18=\left(h-\left(-3\right)\right)\left(h-\left(-6\right)\right)$$

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

$$h^{2}+9h+18=\left(h+3\right)\left(h+6\right)$$

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.

$$x ^ 2 +9x +18 = 0$$

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

$$r + s = -9 rs = 18$$

Two numbers r and s sum up to -9 exactly when the average of the two numbers is \frac{1}{2}*-9 = -\frac{9}{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>

$$r = -\frac{9}{2} - u s = -\frac{9}{2} + u$$

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

$$(-\frac{9}{2} - u) (-\frac{9}{2} + u) = 18$$

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

$$\frac{81}{4} - u^2 = 18$$

Simplify the expression by subtracting \frac{81}{4} on both sides

$$-u^2 = 18-\frac{81}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =-\frac{9}{2} - \frac{3}{2} = -6 s = -\frac{9}{2} + \frac{3}{2} = -3$$

Evaluate

$$\left(h+3\right)\left(h+6\right)$$