Solve 11x^2-98x-120

$$11 x ^ { 2 } - 98 x - 120$$

Factor

$$\left(x-10\right)\left(11x+12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 11x^{2}+ax+bx-120. To find a and b, set up a system to be solved.

$$a+b=-98 ab=11\left(-120\right)=-1320$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1320.

$$1,-1320 2,-660 3,-440 4,-330 5,-264 6,-220 8,-165 10,-132 11,-120 12,-110 15,-88 20,-66 22,-60 24,-55 30,-44 33,-40$$

Calculate the sum for each pair.

$$1-1320=-1319 2-660=-658 3-440=-437 4-330=-326 5-264=-259 6-220=-214 8-165=-157 10-132=-122 11-120=-109 12-110=-98 15-88=-73 20-66=-46 22-60=-38 24-55=-31 30-44=-14 33-40=-7$$

The solution is the pair that gives sum -98.

$$a=-110 b=12$$

Rewrite 11x^{2}-98x-120 as \left(11x^{2}-110x\right)+\left(12x-120\right).

$$\left(11x^{2}-110x\right)+\left(12x-120\right)$$

Factor out 11x in the first and 12 in the second group.

$$11x\left(x-10\right)+12\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(11x+12\right)$$

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.

$$11x^{2}-98x-120=0$$

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.

$$x=\frac{-\left(-98\right)±\sqrt{\left(-98\right)^{2}-4\times 11\left(-120\right)}}{2\times 11}$$

Square -98.

$$x=\frac{-\left(-98\right)±\sqrt{9604-4\times 11\left(-120\right)}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-\left(-98\right)±\sqrt{9604-44\left(-120\right)}}{2\times 11}$$

Multiply -44 times -120.

$$x=\frac{-\left(-98\right)±\sqrt{9604+5280}}{2\times 11}$$

Add 9604 to 5280.

$$x=\frac{-\left(-98\right)±\sqrt{14884}}{2\times 11}$$

Take the square root of 14884.

$$x=\frac{-\left(-98\right)±122}{2\times 11}$$

The opposite of -98 is 98.

$$x=\frac{98±122}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{98±122}{22}$$

Now solve the equation x=\frac{98±122}{22} when ± is plus. Add 98 to 122.

$$x=\frac{220}{22}$$

Divide 220 by 22.

$$x=10$$

Now solve the equation x=\frac{98±122}{22} when ± is minus. Subtract 122 from 98.

$$x=-\frac{24}{22}$$

Reduce the fraction \frac{-24}{22} to lowest terms by extracting and canceling out 2.

$$x=-\frac{12}{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10 for x_{1} and -\frac{12}{11} for x_{2}.

$$11x^{2}-98x-120=11\left(x-10\right)\left(x-\left(-\frac{12}{11}\right)\right)$$

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

$$11x^{2}-98x-120=11\left(x-10\right)\left(x+\frac{12}{11}\right)$$

Add \frac{12}{11} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$11x^{2}-98x-120=11\left(x-10\right)\times \frac{11x+12}{11}$$

Cancel out 11, the greatest common factor in 11 and 11.

$$11x^{2}-98x-120=\left(x-10\right)\left(11x+12\right)$$

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.This is achieved by dividing both sides of the equation by 11

$$x ^ 2 -\frac{98}{11}x -\frac{120}{11} = 0$$

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

$$r + s = \frac{98}{11} rs = -\frac{120}{11}$$

Two numbers r and s sum up to \frac{98}{11} exactly when the average of the two numbers is \frac{1}{2}*\frac{98}{11} = \frac{49}{11}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{49}{11} - u s = \frac{49}{11} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{120}{11}

$$(\frac{49}{11} - u) (\frac{49}{11} + u) = -\frac{120}{11}$$

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

$$\frac{2401}{121} - u^2 = -\frac{120}{11}$$

Simplify the expression by subtracting \frac{2401}{121} on both sides

$$-u^2 = -\frac{120}{11}-\frac{2401}{121} = -\frac{3721}{121}$$

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

$$u^2 = \frac{3721}{121} u = \pm\sqrt{\frac{3721}{121}} = \pm \frac{61}{11}$$

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

$$r =\frac{49}{11} - \frac{61}{11} = -1.091 s = \frac{49}{11} + \frac{61}{11} = 10$$

Evaluate

$$\left(x-10\right)\left(11x+12\right)$$