Solve x^2+121x+120=

$$x ^ { 2 } + 121 x + 120 =$$

Factor

$$\left(x+1\right)\left(x+120\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+120. To find a and b, set up a system to be solved.

$$a+b=121 ab=1\times 120=120$$

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 120.

$$1,120 2,60 3,40 4,30 5,24 6,20 8,15 10,12$$

Calculate the sum for each pair.

$$1+120=121 2+60=62 3+40=43 4+30=34 5+24=29 6+20=26 8+15=23 10+12=22$$

The solution is the pair that gives sum 121.

$$a=1 b=120$$

Rewrite x^{2}+121x+120 as \left(x^{2}+x\right)+\left(120x+120\right).

$$\left(x^{2}+x\right)+\left(120x+120\right)$$

Factor out x in the first and 120 in the second group.

$$x\left(x+1\right)+120\left(x+1\right)$$

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

$$\left(x+1\right)\left(x+120\right)$$

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.

$$x^{2}+121x+120=0$$

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.

$$x=\frac{-121±\sqrt{121^{2}-4\times 120}}{2}$$

Square 121.

$$x=\frac{-121±\sqrt{14641-4\times 120}}{2}$$

Multiply -4 times 120.

$$x=\frac{-121±\sqrt{14641-480}}{2}$$

Add 14641 to -480.

$$x=\frac{-121±\sqrt{14161}}{2}$$

Take the square root of 14161.

$$x=\frac{-121±119}{2}$$

Now solve the equation x=\frac{-121±119}{2} when ± is plus. Add -121 to 119.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-121±119}{2} when ± is minus. Subtract 119 from -121.

$$x=-\frac{240}{2}$$

Divide -240 by 2.

$$x=-120$$

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 -120 for x_{2}.

$$x^{2}+121x+120=\left(x-\left(-1\right)\right)\left(x-\left(-120\right)\right)$$

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

$$x^{2}+121x+120=\left(x+1\right)\left(x+120\right)$$

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.

$$x ^ 2 +121x +120 = 0$$

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

$$r + s = -121 rs = 120$$

Two numbers r and s sum up to -121 exactly when the average of the two numbers is \frac{1}{2}*-121 = -\frac{121}{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>

$$r = -\frac{121}{2} - u s = -\frac{121}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 120

$$(-\frac{121}{2} - u) (-\frac{121}{2} + u) = 120$$

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

$$\frac{14641}{4} - u^2 = 120$$

Simplify the expression by subtracting \frac{14641}{4} on both sides

$$-u^2 = 120-\frac{14641}{4} = -\frac{14161}{4}$$

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

$$u^2 = \frac{14161}{4} u = \pm\sqrt{\frac{14161}{4}} = \pm \frac{119}{2}$$

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

$$r =-\frac{121}{2} - \frac{119}{2} = -120 s = -\frac{121}{2} + \frac{119}{2} = -1$$

Evaluate

$$\left(x+1\right)\left(x+120\right)$$