Solve 6x^2-11x-35

$$6 { x }^{ 2 } -11x-35$$

Factor

$$\left(2x-7\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-11 ab=6\left(-35\right)=-210$$

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 -210.

$$1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15$$

Calculate the sum for each pair.

$$1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1$$

The solution is the pair that gives sum -11.

$$a=-21 b=10$$

Rewrite 6x^{2}-11x-35 as \left(6x^{2}-21x\right)+\left(10x-35\right).

$$\left(6x^{2}-21x\right)+\left(10x-35\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(2x-7\right)+5\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(3x+5\right)$$

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.

$$6x^{2}-11x-35=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-35\right)}}{2\times 6}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-35\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-24\left(-35\right)}}{2\times 6}$$

Multiply -24 times -35.

$$x=\frac{-\left(-11\right)±\sqrt{121+840}}{2\times 6}$$

Add 121 to 840.

$$x=\frac{-\left(-11\right)±\sqrt{961}}{2\times 6}$$

Take the square root of 961.

$$x=\frac{-\left(-11\right)±31}{2\times 6}$$

The opposite of -11 is 11.

$$x=\frac{11±31}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{11±31}{12}$$

Now solve the equation x=\frac{11±31}{12} when ± is plus. Add 11 to 31.

$$x=\frac{42}{12}$$

Reduce the fraction \frac{42}{12} to lowest terms by extracting and canceling out 6.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{11±31}{12} when ± is minus. Subtract 31 from 11.

$$x=-\frac{20}{12}$$

Reduce the fraction \frac{-20}{12} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{2} for x_{1} and -\frac{5}{3} for x_{2}.

$$6x^{2}-11x-35=6\left(x-\frac{7}{2}\right)\left(x-\left(-\frac{5}{3}\right)\right)$$

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

$$6x^{2}-11x-35=6\left(x-\frac{7}{2}\right)\left(x+\frac{5}{3}\right)$$

Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x-35=6\times \frac{2x-7}{2}\left(x+\frac{5}{3}\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x-35=6\times \frac{2x-7}{2}\times \frac{3x+5}{3}$$

Multiply \frac{2x-7}{2} times \frac{3x+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-11x-35=6\times \frac{\left(2x-7\right)\left(3x+5\right)}{2\times 3}$$

Multiply 2 times 3.

$$6x^{2}-11x-35=6\times \frac{\left(2x-7\right)\left(3x+5\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6x^{2}-11x-35=\left(2x-7\right)\left(3x+5\right)$$

Evaluate

$$\left(2x-7\right)\left(3x+5\right)$$