Solve 3x^2-49x-1700

$$3 { x }^{ 2 } -49x-1700$$

Factor

$$\left(3x-100\right)\left(x+17\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-1700. To find a and b, set up a system to be solved.

$$a+b=-49 ab=3\left(-1700\right)=-5100$$

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 -5100.

$$1,-5100 2,-2550 3,-1700 4,-1275 5,-1020 6,-850 10,-510 12,-425 15,-340 17,-300 20,-255 25,-204 30,-170 34,-150 50,-102 51,-100 60,-85 68,-75$$

Calculate the sum for each pair.

$$1-5100=-5099 2-2550=-2548 3-1700=-1697 4-1275=-1271 5-1020=-1015 6-850=-844 10-510=-500 12-425=-413 15-340=-325 17-300=-283 20-255=-235 25-204=-179 30-170=-140 34-150=-116 50-102=-52 51-100=-49 60-85=-25 68-75=-7$$

The solution is the pair that gives sum -49.

$$a=-100 b=51$$

Rewrite 3x^{2}-49x-1700 as \left(3x^{2}-100x\right)+\left(51x-1700\right).

$$\left(3x^{2}-100x\right)+\left(51x-1700\right)$$

Factor out x in the first and 17 in the second group.

$$x\left(3x-100\right)+17\left(3x-100\right)$$

Factor out common term 3x-100 by using distributive property.

$$\left(3x-100\right)\left(x+17\right)$$

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.

$$3x^{2}-49x-1700=0$$

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.

$$x=\frac{-\left(-49\right)±\sqrt{\left(-49\right)^{2}-4\times 3\left(-1700\right)}}{2\times 3}$$

Square -49.

$$x=\frac{-\left(-49\right)±\sqrt{2401-4\times 3\left(-1700\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-49\right)±\sqrt{2401-12\left(-1700\right)}}{2\times 3}$$

Multiply -12 times -1700.

$$x=\frac{-\left(-49\right)±\sqrt{2401+20400}}{2\times 3}$$

Add 2401 to 20400.

$$x=\frac{-\left(-49\right)±\sqrt{22801}}{2\times 3}$$

Take the square root of 22801.

$$x=\frac{-\left(-49\right)±151}{2\times 3}$$

The opposite of -49 is 49.

$$x=\frac{49±151}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{49±151}{6}$$

Now solve the equation x=\frac{49±151}{6} when ± is plus. Add 49 to 151.

$$x=\frac{200}{6}$$

Reduce the fraction \frac{200}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{100}{3}$$

Now solve the equation x=\frac{49±151}{6} when ± is minus. Subtract 151 from 49.

$$x=-\frac{102}{6}$$

Divide -102 by 6.

$$x=-17$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{100}{3} for x_{1} and -17 for x_{2}.

$$3x^{2}-49x-1700=3\left(x-\frac{100}{3}\right)\left(x-\left(-17\right)\right)$$

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

$$3x^{2}-49x-1700=3\left(x-\frac{100}{3}\right)\left(x+17\right)$$

Subtract \frac{100}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}-49x-1700=3\times \frac{3x-100}{3}\left(x+17\right)$$

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

$$3x^{2}-49x-1700=\left(3x-100\right)\left(x+17\right)$$

Evaluate

$$\left(3x-100\right)\left(x+17\right)$$