Solve 5y^2+2y-3

$$5 y ^ { 2 } + 2 y - 3$$

Factor

$$\left(5y-3\right)\left(y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=5\left(-3\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 5y^{2}+2y-3 as \left(5y^{2}-3y\right)+\left(5y-3\right).

$$\left(5y^{2}-3y\right)+\left(5y-3\right)$$

Factor out y in 5y^{2}-3y.

$$y\left(5y-3\right)+5y-3$$

Factor out common term 5y-3 by using distributive property.

$$\left(5y-3\right)\left(y+1\right)$$

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.

$$5y^{2}+2y-3=0$$

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.

$$y=\frac{-2±\sqrt{2^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Square 2.

$$y=\frac{-2±\sqrt{4-4\times 5\left(-3\right)}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-2±\sqrt{4-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$y=\frac{-2±\sqrt{4+60}}{2\times 5}$$

Add 4 to 60.

$$y=\frac{-2±\sqrt{64}}{2\times 5}$$

Take the square root of 64.

$$y=\frac{-2±8}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-2±8}{10}$$

Now solve the equation y=\frac{-2±8}{10} when ± is plus. Add -2 to 8.

$$y=\frac{6}{10}$$

Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.

$$y=\frac{3}{5}$$

Now solve the equation y=\frac{-2±8}{10} when ± is minus. Subtract 8 from -2.

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

Divide -10 by 10.

$$y=-1$$

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

$$5y^{2}+2y-3=5\left(y-\frac{3}{5}\right)\left(y-\left(-1\right)\right)$$

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

$$5y^{2}+2y-3=5\left(y-\frac{3}{5}\right)\left(y+1\right)$$

Subtract \frac{3}{5} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5y^{2}+2y-3=5\times \frac{5y-3}{5}\left(y+1\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5y^{2}+2y-3=\left(5y-3\right)\left(y+1\right)$$

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 5

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

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

$$r + s = -\frac{2}{5} rs = -\frac{3}{5}$$

Two numbers r and s sum up to -\frac{2}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{5} = -\frac{1}{5}. 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>

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

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

$$(-\frac{1}{5} - u) (-\frac{1}{5} + u) = -\frac{3}{5}$$

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

$$\frac{1}{25} - u^2 = -\frac{3}{5}$$

Simplify the expression by subtracting \frac{1}{25} on both sides

$$-u^2 = -\frac{3}{5}-\frac{1}{25} = -\frac{16}{25}$$

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

$$u^2 = \frac{16}{25} u = \pm\sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$

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

$$r =-\frac{1}{5} - \frac{4}{5} = -1 s = -\frac{1}{5} + \frac{4}{5} = 0.600$$

Evaluate

$$\left(5y-3\right)\left(y+1\right)$$