Solve 6y^2+y-5

$$6 y ^ { 2 } + y - 5$$

Factor

$$\left(6y-5\right)\left(y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=1 ab=6\left(-5\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=-5 b=6$$

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

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

Factor out y in 6y^{2}-5y.

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

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

$$\left(6y-5\right)\left(y+1\right)$$

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.

$$6y^{2}+y-5=0$$

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.

$$y=\frac{-1±\sqrt{1^{2}-4\times 6\left(-5\right)}}{2\times 6}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 6\left(-5\right)}}{2\times 6}$$

Multiply -4 times 6.

$$y=\frac{-1±\sqrt{1-24\left(-5\right)}}{2\times 6}$$

Multiply -24 times -5.

$$y=\frac{-1±\sqrt{1+120}}{2\times 6}$$

Add 1 to 120.

$$y=\frac{-1±\sqrt{121}}{2\times 6}$$

Take the square root of 121.

$$y=\frac{-1±11}{2\times 6}$$

Multiply 2 times 6.

$$y=\frac{-1±11}{12}$$

Now solve the equation y=\frac{-1±11}{12} when ± is plus. Add -1 to 11.

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

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{-1±11}{12} when ± is minus. Subtract 11 from -1.

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

Divide -12 by 12.

$$y=-1$$

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 -1 for x_{2}.

$$6y^{2}+y-5=6\left(y-\frac{5}{6}\right)\left(y-\left(-1\right)\right)$$

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

$$6y^{2}+y-5=6\left(y-\frac{5}{6}\right)\left(y+1\right)$$

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

$$6y^{2}+y-5=6\times \frac{6y-5}{6}\left(y+1\right)$$

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

$$6y^{2}+y-5=\left(6y-5\right)\left(y+1\right)$$

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

$$x ^ 2 +\frac{1}{6}x -\frac{5}{6} = 0$$

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

$$r + s = -\frac{1}{6} rs = -\frac{5}{6}$$

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

$$r = -\frac{1}{12} - u s = -\frac{1}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{6}

$$(-\frac{1}{12} - u) (-\frac{1}{12} + u) = -\frac{5}{6}$$

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

$$\frac{1}{144} - u^2 = -\frac{5}{6}$$

Simplify the expression by subtracting \frac{1}{144} on both sides

$$-u^2 = -\frac{5}{6}-\frac{1}{144} = -\frac{121}{144}$$

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

$$u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}$$

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

$$r =-\frac{1}{12} - \frac{11}{12} = -1 s = -\frac{1}{12} + \frac{11}{12} = 0.833$$

Evaluate

$$\left(6y-5\right)\left(y+1\right)$$