Solve 6a^2-11a-10

$$6 a ^ { 2 } - 11 a - 10$$

Factor

$$\left(2a-5\right)\left(3a+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-10. To find p and q, set up a system to be solved.

$$p+q=-11 pq=6\left(-10\right)=-60$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$$1,-60 2,-30 3,-20 4,-15 5,-12 6,-10$$

Calculate the sum for each pair.

$$1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4$$

The solution is the pair that gives sum -11.

$$p=-15 q=4$$

Rewrite 6a^{2}-11a-10 as \left(6a^{2}-15a\right)+\left(4a-10\right).

$$\left(6a^{2}-15a\right)+\left(4a-10\right)$$

Factor out 3a in the first and 2 in the second group.

$$3a\left(2a-5\right)+2\left(2a-5\right)$$

Factor out common term 2a-5 by using distributive property.

$$\left(2a-5\right)\left(3a+2\right)$$

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.

$$6a^{2}-11a-10=0$$

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.

$$a=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-10\right)}}{2\times 6}$$

Square -11.

$$a=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-10\right)}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-\left(-11\right)±\sqrt{121-24\left(-10\right)}}{2\times 6}$$

Multiply -24 times -10.

$$a=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 6}$$

Add 121 to 240.

$$a=\frac{-\left(-11\right)±\sqrt{361}}{2\times 6}$$

Take the square root of 361.

$$a=\frac{-\left(-11\right)±19}{2\times 6}$$

The opposite of -11 is 11.

$$a=\frac{11±19}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{11±19}{12}$$

Now solve the equation a=\frac{11±19}{12} when ± is plus. Add 11 to 19.

$$a=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

$$a=\frac{5}{2}$$

Now solve the equation a=\frac{11±19}{12} when ± is minus. Subtract 19 from 11.

$$a=-\frac{8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

$$a=-\frac{2}{3}$$

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

$$6a^{2}-11a-10=6\left(a-\frac{5}{2}\right)\left(a-\left(-\frac{2}{3}\right)\right)$$

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

$$6a^{2}-11a-10=6\left(a-\frac{5}{2}\right)\left(a+\frac{2}{3}\right)$$

Subtract \frac{5}{2} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6a^{2}-11a-10=6\times \frac{2a-5}{2}\left(a+\frac{2}{3}\right)$$

Add \frac{2}{3} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6a^{2}-11a-10=6\times \frac{2a-5}{2}\times \frac{3a+2}{3}$$

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

$$6a^{2}-11a-10=6\times \frac{\left(2a-5\right)\left(3a+2\right)}{2\times 3}$$

Multiply 2 times 3.

$$6a^{2}-11a-10=6\times \frac{\left(2a-5\right)\left(3a+2\right)}{6}$$

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

$$6a^{2}-11a-10=\left(2a-5\right)\left(3a+2\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 6

$$x ^ 2 -\frac{11}{6}x -\frac{5}{3} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{11}{6} rs = -\frac{5}{3}$$

Two numbers r and s sum up to \frac{11}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{6} = \frac{11}{12}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{11}{12} - u s = \frac{11}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(\frac{11}{12} - u) (\frac{11}{12} + u) = -\frac{5}{3}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{121}{144} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{121}{144} on both sides

$$-u^2 = -\frac{5}{3}-\frac{121}{144} = -\frac{361}{144}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{11}{12} - \frac{19}{12} = -0.667 s = \frac{11}{12} + \frac{19}{12} = 2.500$$

Evaluate

$$\left(2a-5\right)\left(3a+2\right)$$