Solve 6y^2+7y-10

$$6 y ^ { 2 } + 7 y - 10$$

Factor

$$\left(6y-5\right)\left(y+2\right)$$

Solution Steps

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

$$a+b=7 ab=6\left(-10\right)=-60$$

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 -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 7.

$$a=-5 b=12$$

Rewrite 6y^{2}+7y-10 as \left(6y^{2}-5y\right)+\left(12y-10\right).

$$\left(6y^{2}-5y\right)+\left(12y-10\right)$$

Factor out y in the first and 2 in the second group.

$$y\left(6y-5\right)+2\left(6y-5\right)$$

Factor out common term 6y-5 by using distributive property.

$$\left(6y-5\right)\left(y+2\right)$$

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.

$$6y^{2}+7y-10=0$$

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.

$$y=\frac{-7±\sqrt{7^{2}-4\times 6\left(-10\right)}}{2\times 6}$$

Square 7.

$$y=\frac{-7±\sqrt{49-4\times 6\left(-10\right)}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-7±\sqrt{49-24\left(-10\right)}}{2\times 6}$$

Multiply -24 times -10.

$$y=\frac{-7±\sqrt{49+240}}{2\times 6}$$

Add 49 to 240.

$$y=\frac{-7±\sqrt{289}}{2\times 6}$$

Take the square root of 289.

$$y=\frac{-7±17}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-7±17}{12}$$

Now solve the equation y=\frac{-7±17}{12} when ± is plus. Add -7 to 17.

$$y=\frac{10}{12}$$

Reduce the fraction \frac{10}{12} to lowest terms by extracting and canceling out 2.

$$y=\frac{5}{6}$$

Now solve the equation y=\frac{-7±17}{12} when ± is minus. Subtract 17 from -7.

$$y=-\frac{24}{12}$$

Divide -24 by 12.

$$y=-2$$

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

$$6y^{2}+7y-10=6\left(y-\frac{5}{6}\right)\left(y-\left(-2\right)\right)$$

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

$$6y^{2}+7y-10=6\left(y-\frac{5}{6}\right)\left(y+2\right)$$

Subtract \frac{5}{6} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6y^{2}+7y-10=6\times \frac{6y-5}{6}\left(y+2\right)$$

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

$$6y^{2}+7y-10=\left(6y-5\right)\left(y+2\right)$$

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

$$x ^ 2 +\frac{7}{6}x -\frac{5}{3} = 0$$

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

$$r + s = -\frac{7}{6} rs = -\frac{5}{3}$$

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{5}{3}

$$(-\frac{7}{12} - u) (-\frac{7}{12} + u) = -\frac{5}{3}$$

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

$$\frac{49}{144} - u^2 = -\frac{5}{3}$$

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

$$-u^2 = -\frac{5}{3}-\frac{49}{144} = -\frac{289}{144}$$

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

$$u^2 = \frac{289}{144} u = \pm\sqrt{\frac{289}{144}} = \pm \frac{17}{12}$$

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

$$r =-\frac{7}{12} - \frac{17}{12} = -2 s = -\frac{7}{12} + \frac{17}{12} = 0.833$$

Evaluate

$$\left(6y-5\right)\left(y+2\right)$$