Solve 11x^2+140x-196=

$$11 { x }^{ 2 } +140x-196=$$

Factor

$$\left(11x-14\right)\left(x+14\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 11x^{2}+ax+bx-196. To find a and b, set up a system to be solved.

$$a+b=140 ab=11\left(-196\right)=-2156$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -2156.

$$-1,2156 -2,1078 -4,539 -7,308 -11,196 -14,154 -22,98 -28,77 -44,49$$

Calculate the sum for each pair.

$$-1+2156=2155 -2+1078=1076 -4+539=535 -7+308=301 -11+196=185 -14+154=140 -22+98=76 -28+77=49 -44+49=5$$

The solution is the pair that gives sum 140.

$$a=-14 b=154$$

Rewrite 11x^{2}+140x-196 as \left(11x^{2}-14x\right)+\left(154x-196\right).

$$\left(11x^{2}-14x\right)+\left(154x-196\right)$$

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

$$x\left(11x-14\right)+14\left(11x-14\right)$$

Factor out common term 11x-14 by using distributive property.

$$\left(11x-14\right)\left(x+14\right)$$

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.

$$11x^{2}+140x-196=0$$

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.

$$x=\frac{-140±\sqrt{140^{2}-4\times 11\left(-196\right)}}{2\times 11}$$

Square 140.

$$x=\frac{-140±\sqrt{19600-4\times 11\left(-196\right)}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-140±\sqrt{19600-44\left(-196\right)}}{2\times 11}$$

Multiply -44 times -196.

$$x=\frac{-140±\sqrt{19600+8624}}{2\times 11}$$

Add 19600 to 8624.

$$x=\frac{-140±\sqrt{28224}}{2\times 11}$$

Take the square root of 28224.

$$x=\frac{-140±168}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{-140±168}{22}$$

Now solve the equation x=\frac{-140±168}{22} when ± is plus. Add -140 to 168.

$$x=\frac{28}{22}$$

Reduce the fraction \frac{28}{22} to lowest terms by extracting and canceling out 2.

$$x=\frac{14}{11}$$

Now solve the equation x=\frac{-140±168}{22} when ± is minus. Subtract 168 from -140.

$$x=-\frac{308}{22}$$

Divide -308 by 22.

$$x=-14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{14}{11} for x_{1} and -14 for x_{2}.

$$11x^{2}+140x-196=11\left(x-\frac{14}{11}\right)\left(x-\left(-14\right)\right)$$

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

$$11x^{2}+140x-196=11\left(x-\frac{14}{11}\right)\left(x+14\right)$$

Subtract \frac{14}{11} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$11x^{2}+140x-196=11\times \frac{11x-14}{11}\left(x+14\right)$$

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

$$11x^{2}+140x-196=\left(11x-14\right)\left(x+14\right)$$

Evaluate

$$\left(11x-14\right)\left(x+14\right)$$