Solve 6x^2-x-145

$$6 { x }^{ 2 } -x-145$$

Factor

$$\left(x-5\right)\left(6x+29\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6x^{2}+ax+bx-145. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-145\right)=-870$$

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 -870.

$$1,-870 2,-435 3,-290 5,-174 6,-145 10,-87 15,-58 29,-30$$

Calculate the sum for each pair.

$$1-870=-869 2-435=-433 3-290=-287 5-174=-169 6-145=-139 10-87=-77 15-58=-43 29-30=-1$$

The solution is the pair that gives sum -1.

$$a=-30 b=29$$

Rewrite 6x^{2}-x-145 as \left(6x^{2}-30x\right)+\left(29x-145\right).

$$\left(6x^{2}-30x\right)+\left(29x-145\right)$$

Factor out 6x in the first and 29 in the second group.

$$6x\left(x-5\right)+29\left(x-5\right)$$

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

$$\left(x-5\right)\left(6x+29\right)$$

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.

$$6x^{2}-x-145=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-145\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-145\right)}}{2\times 6}$$

Multiply -24 times -145.

$$x=\frac{-\left(-1\right)±\sqrt{1+3480}}{2\times 6}$$

Add 1 to 3480.

$$x=\frac{-\left(-1\right)±\sqrt{3481}}{2\times 6}$$

Take the square root of 3481.

$$x=\frac{-\left(-1\right)±59}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±59}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±59}{12}$$

Now solve the equation x=\frac{1±59}{12} when ± is plus. Add 1 to 59.

$$x=\frac{60}{12}$$

Divide 60 by 12.

$$x=5$$

Now solve the equation x=\frac{1±59}{12} when ± is minus. Subtract 59 from 1.

$$x=-\frac{58}{12}$$

Reduce the fraction \frac{-58}{12} to lowest terms by extracting and canceling out 2.

$$x=-\frac{29}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -\frac{29}{6} for x_{2}.

$$6x^{2}-x-145=6\left(x-5\right)\left(x-\left(-\frac{29}{6}\right)\right)$$

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

$$6x^{2}-x-145=6\left(x-5\right)\left(x+\frac{29}{6}\right)$$

Add \frac{29}{6} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6x^{2}-x-145=6\left(x-5\right)\times \frac{6x+29}{6}$$

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

$$6x^{2}-x-145=\left(x-5\right)\left(6x+29\right)$$

Evaluate

$$\left(x-5\right)\left(6x+29\right)$$