Solve 3x^2-119x+1136

$$3 { x }^{ 2 } -119x+1136$$

Factor

$$\left(x-16\right)\left(3x-71\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx+1136. To find a and b, set up a system to be solved.

$$a+b=-119 ab=3\times 1136=3408$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 3408.

$$-1,-3408 -2,-1704 -3,-1136 -4,-852 -6,-568 -8,-426 -12,-284 -16,-213 -24,-142 -48,-71$$

Calculate the sum for each pair.

$$-1-3408=-3409 -2-1704=-1706 -3-1136=-1139 -4-852=-856 -6-568=-574 -8-426=-434 -12-284=-296 -16-213=-229 -24-142=-166 -48-71=-119$$

The solution is the pair that gives sum -119.

$$a=-71 b=-48$$

Rewrite 3x^{2}-119x+1136 as \left(3x^{2}-71x\right)+\left(-48x+1136\right).

$$\left(3x^{2}-71x\right)+\left(-48x+1136\right)$$

Factor out x in the first and -16 in the second group.

$$x\left(3x-71\right)-16\left(3x-71\right)$$

Factor out common term 3x-71 by using distributive property.

$$\left(3x-71\right)\left(x-16\right)$$

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.

$$3x^{2}-119x+1136=0$$

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.

$$x=\frac{-\left(-119\right)±\sqrt{\left(-119\right)^{2}-4\times 3\times 1136}}{2\times 3}$$

Square -119.

$$x=\frac{-\left(-119\right)±\sqrt{14161-4\times 3\times 1136}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-119\right)±\sqrt{14161-12\times 1136}}{2\times 3}$$

Multiply -12 times 1136.

$$x=\frac{-\left(-119\right)±\sqrt{14161-13632}}{2\times 3}$$

Add 14161 to -13632.

$$x=\frac{-\left(-119\right)±\sqrt{529}}{2\times 3}$$

Take the square root of 529.

$$x=\frac{-\left(-119\right)±23}{2\times 3}$$

The opposite of -119 is 119.

$$x=\frac{119±23}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{119±23}{6}$$

Now solve the equation x=\frac{119±23}{6} when ± is plus. Add 119 to 23.

$$x=\frac{142}{6}$$

Reduce the fraction \frac{142}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{71}{3}$$

Now solve the equation x=\frac{119±23}{6} when ± is minus. Subtract 23 from 119.

$$x=\frac{96}{6}$$

Divide 96 by 6.

$$x=16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{71}{3} for x_{1} and 16 for x_{2}.

$$3x^{2}-119x+1136=3\left(x-\frac{71}{3}\right)\left(x-16\right)$$

Subtract \frac{71}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-119x+1136=3\times \frac{3x-71}{3}\left(x-16\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}-119x+1136=\left(3x-71\right)\left(x-16\right)$$

Evaluate

$$\left(x-16\right)\left(3x-71\right)$$