Solve n^2+21n+98

$$n ^ { 2 } + 21 n + 98$$

Factor

$$\left(n+7\right)\left(n+14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as n^{2}+an+bn+98. To find a and b, set up a system to be solved.

$$a+b=21 ab=1\times 98=98$$

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 98.

$$1,98 2,49 7,14$$

Calculate the sum for each pair.

$$1+98=99 2+49=51 7+14=21$$

The solution is the pair that gives sum 21.

$$a=7 b=14$$

Rewrite n^{2}+21n+98 as \left(n^{2}+7n\right)+\left(14n+98\right).

$$\left(n^{2}+7n\right)+\left(14n+98\right)$$

Factor out n in the first and 14 in the second group.

$$n\left(n+7\right)+14\left(n+7\right)$$

Factor out common term n+7 by using distributive property.

$$\left(n+7\right)\left(n+14\right)$$

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.

$$n^{2}+21n+98=0$$

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.

$$n=\frac{-21±\sqrt{21^{2}-4\times 98}}{2}$$

Square 21.

$$n=\frac{-21±\sqrt{441-4\times 98}}{2}$$

Multiply -4 times 98.

$$n=\frac{-21±\sqrt{441-392}}{2}$$

Add 441 to -392.

$$n=\frac{-21±\sqrt{49}}{2}$$

Take the square root of 49.

$$n=\frac{-21±7}{2}$$

Now solve the equation n=\frac{-21±7}{2} when ± is plus. Add -21 to 7.

$$n=-\frac{14}{2}$$

Divide -14 by 2.

$$n=-7$$

Now solve the equation n=\frac{-21±7}{2} when ± is minus. Subtract 7 from -21.

$$n=-\frac{28}{2}$$

Divide -28 by 2.

$$n=-14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -7 for x_{1} and -14 for x_{2}.

$$n^{2}+21n+98=\left(n-\left(-7\right)\right)\left(n-\left(-14\right)\right)$$

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

$$n^{2}+21n+98=\left(n+7\right)\left(n+14\right)$$

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.

$$x ^ 2 +21x +98 = 0$$

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

$$r + s = -21 rs = 98$$

Two numbers r and s sum up to -21 exactly when the average of the two numbers is \frac{1}{2}*-21 = -\frac{21}{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>

$$r = -\frac{21}{2} - u s = -\frac{21}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 98

$$(-\frac{21}{2} - u) (-\frac{21}{2} + u) = 98$$

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

$$\frac{441}{4} - u^2 = 98$$

Simplify the expression by subtracting \frac{441}{4} on both sides

$$-u^2 = 98-\frac{441}{4} = -\frac{49}{4}$$

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

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

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

$$r =-\frac{21}{2} - \frac{7}{2} = -14 s = -\frac{21}{2} + \frac{7}{2} = -7$$

Evaluate

$$\left(n+7\right)\left(n+14\right)$$