Solve 10x^2+13x-231

$$10 { x }^{ 2 } +13x-231$$

Factor

$$\left(5x-21\right)\left(2x+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-231. To find a and b, set up a system to be solved.

$$a+b=13 ab=10\left(-231\right)=-2310$$

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 -2310.

$$-1,2310 -2,1155 -3,770 -5,462 -6,385 -7,330 -10,231 -11,210 -14,165 -15,154 -21,110 -22,105 -30,77 -33,70 -35,66 -42,55$$

Calculate the sum for each pair.

$$-1+2310=2309 -2+1155=1153 -3+770=767 -5+462=457 -6+385=379 -7+330=323 -10+231=221 -11+210=199 -14+165=151 -15+154=139 -21+110=89 -22+105=83 -30+77=47 -33+70=37 -35+66=31 -42+55=13$$

The solution is the pair that gives sum 13.

$$a=-42 b=55$$

Rewrite 10x^{2}+13x-231 as \left(10x^{2}-42x\right)+\left(55x-231\right).

$$\left(10x^{2}-42x\right)+\left(55x-231\right)$$

Factor out 2x in the first and 11 in the second group.

$$2x\left(5x-21\right)+11\left(5x-21\right)$$

Factor out common term 5x-21 by using distributive property.

$$\left(5x-21\right)\left(2x+11\right)$$

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.

$$10x^{2}+13x-231=0$$

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.

$$x=\frac{-13±\sqrt{13^{2}-4\times 10\left(-231\right)}}{2\times 10}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 10\left(-231\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-13±\sqrt{169-40\left(-231\right)}}{2\times 10}$$

Multiply -40 times -231.

$$x=\frac{-13±\sqrt{169+9240}}{2\times 10}$$

Add 169 to 9240.

$$x=\frac{-13±\sqrt{9409}}{2\times 10}$$

Take the square root of 9409.

$$x=\frac{-13±97}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-13±97}{20}$$

Now solve the equation x=\frac{-13±97}{20} when ± is plus. Add -13 to 97.

$$x=\frac{84}{20}$$

Reduce the fraction \frac{84}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{21}{5}$$

Now solve the equation x=\frac{-13±97}{20} when ± is minus. Subtract 97 from -13.

$$x=-\frac{110}{20}$$

Reduce the fraction \frac{-110}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{11}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{21}{5} for x_{1} and -\frac{11}{2} for x_{2}.

$$10x^{2}+13x-231=10\left(x-\frac{21}{5}\right)\left(x-\left(-\frac{11}{2}\right)\right)$$

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

$$10x^{2}+13x-231=10\left(x-\frac{21}{5}\right)\left(x+\frac{11}{2}\right)$$

Subtract \frac{21}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+13x-231=10\times \frac{5x-21}{5}\left(x+\frac{11}{2}\right)$$

Add \frac{11}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+13x-231=10\times \frac{5x-21}{5}\times \frac{2x+11}{2}$$

Multiply \frac{5x-21}{5} times \frac{2x+11}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+13x-231=10\times \frac{\left(5x-21\right)\left(2x+11\right)}{5\times 2}$$

Multiply 5 times 2.

$$10x^{2}+13x-231=10\times \frac{\left(5x-21\right)\left(2x+11\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10x^{2}+13x-231=\left(5x-21\right)\left(2x+11\right)$$

Evaluate

$$\left(5x-21\right)\left(2x+11\right)$$