Solve 6a^2+7a-20

$$6 a ^ { 2 } + 7 a - 20$$

Factor

$$\left(3a-4\right)\left(2a+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6a^{2}+pa+qa-20. To find p and q, set up a system to be solved.

$$p+q=7 pq=6\left(-20\right)=-120$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -120.

$$-1,120 -2,60 -3,40 -4,30 -5,24 -6,20 -8,15 -10,12$$

Calculate the sum for each pair.

$$-1+120=119 -2+60=58 -3+40=37 -4+30=26 -5+24=19 -6+20=14 -8+15=7 -10+12=2$$

The solution is the pair that gives sum 7.

$$p=-8 q=15$$

Rewrite 6a^{2}+7a-20 as \left(6a^{2}-8a\right)+\left(15a-20\right).

$$\left(6a^{2}-8a\right)+\left(15a-20\right)$$

Factor out 2a in the first and 5 in the second group.

$$2a\left(3a-4\right)+5\left(3a-4\right)$$

Factor out common term 3a-4 by using distributive property.

$$\left(3a-4\right)\left(2a+5\right)$$

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.

$$6a^{2}+7a-20=0$$

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.

$$a=\frac{-7±\sqrt{7^{2}-4\times 6\left(-20\right)}}{2\times 6}$$

Square 7.

$$a=\frac{-7±\sqrt{49-4\times 6\left(-20\right)}}{2\times 6}$$

Multiply -4 times 6.

$$a=\frac{-7±\sqrt{49-24\left(-20\right)}}{2\times 6}$$

Multiply -24 times -20.

$$a=\frac{-7±\sqrt{49+480}}{2\times 6}$$

Add 49 to 480.

$$a=\frac{-7±\sqrt{529}}{2\times 6}$$

Take the square root of 529.

$$a=\frac{-7±23}{2\times 6}$$

Multiply 2 times 6.

$$a=\frac{-7±23}{12}$$

Now solve the equation a=\frac{-7±23}{12} when ± is plus. Add -7 to 23.

$$a=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$a=\frac{4}{3}$$

Now solve the equation a=\frac{-7±23}{12} when ± is minus. Subtract 23 from -7.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -\frac{5}{2} for x_{2}.

$$6a^{2}+7a-20=6\left(a-\frac{4}{3}\right)\left(a-\left(-\frac{5}{2}\right)\right)$$

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

$$6a^{2}+7a-20=6\left(a-\frac{4}{3}\right)\left(a+\frac{5}{2}\right)$$

Subtract \frac{4}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6a^{2}+7a-20=6\times \frac{3a-4}{3}\left(a+\frac{5}{2}\right)$$

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

$$6a^{2}+7a-20=6\times \frac{3a-4}{3}\times \frac{2a+5}{2}$$

Multiply \frac{3a-4}{3} times \frac{2a+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6a^{2}+7a-20=6\times \frac{\left(3a-4\right)\left(2a+5\right)}{3\times 2}$$

Multiply 3 times 2.

$$6a^{2}+7a-20=6\times \frac{\left(3a-4\right)\left(2a+5\right)}{6}$$

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

$$6a^{2}+7a-20=\left(3a-4\right)\left(2a+5\right)$$

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

$$x ^ 2 +\frac{7}{6}x -\frac{10}{3} = 0$$

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

$$r + s = -\frac{7}{6} rs = -\frac{10}{3}$$

Two numbers r and s sum up to -\frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{6} = -\frac{7}{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>

$$r = -\frac{7}{12} - u s = -\frac{7}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{10}{3}

$$(-\frac{7}{12} - u) (-\frac{7}{12} + u) = -\frac{10}{3}$$

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

$$\frac{49}{144} - u^2 = -\frac{10}{3}$$

Simplify the expression by subtracting \frac{49}{144} on both sides

$$-u^2 = -\frac{10}{3}-\frac{49}{144} = -\frac{529}{144}$$

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

$$u^2 = \frac{529}{144} u = \pm\sqrt{\frac{529}{144}} = \pm \frac{23}{12}$$

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

$$r =-\frac{7}{12} - \frac{23}{12} = -2.500 s = -\frac{7}{12} + \frac{23}{12} = 1.333$$

Evaluate

$$\left(3a-4\right)\left(2a+5\right)$$