Solve -5y^2-8y+4

$$- 5 y ^ { 2 } - 8 y + 4$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -5y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-5\times 4=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=2 b=-10$$

Rewrite -5y^{2}-8y+4 as \left(-5y^{2}+2y\right)+\left(-10y+4\right).

$$\left(-5y^{2}+2y\right)+\left(-10y+4\right)$$

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

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

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

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

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.

$$-5y^{2}-8y+4=0$$

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.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\left(-5\right)\times 4}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-\left(-8\right)±\sqrt{64+20\times 4}}{2\left(-5\right)}$$

Multiply 20 times 4.

$$y=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-5\right)}$$

Add 64 to 80.

$$y=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-5\right)}$$

Take the square root of 144.

$$y=\frac{-\left(-8\right)±12}{2\left(-5\right)}$$

The opposite of -8 is 8.

$$y=\frac{8±12}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{8±12}{-10}$$

Now solve the equation y=\frac{8±12}{-10} when ± is plus. Add 8 to 12.

$$y=\frac{20}{-10}$$

Divide 20 by -10.

$$y=-2$$

Now solve the equation y=\frac{8±12}{-10} when ± is minus. Subtract 12 from 8.

$$y=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

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

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

$$-5y^{2}-8y+4=-5\left(y-\left(-2\right)\right)\left(y-\frac{2}{5}\right)$$

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

$$-5y^{2}-8y+4=-5\left(y+2\right)\left(y-\frac{2}{5}\right)$$

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

$$-5y^{2}-8y+4=-5\left(y+2\right)\times \frac{-5y+2}{-5}$$

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

$$-5y^{2}-8y+4=\left(y+2\right)\left(-5y+2\right)$$

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.

$$x ^ 2 +\frac{8}{5}x -\frac{4}{5} = 0$$

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

$$r + s = -\frac{8}{5} rs = -\frac{4}{5}$$

Two numbers r and s sum up to -\frac{8}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{5} = -\frac{4}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}

$$(-\frac{4}{5} - u) (-\frac{4}{5} + u) = -\frac{4}{5}$$

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

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

Simplify the expression by subtracting \frac{16}{25} on both sides

$$-u^2 = -\frac{4}{5}-\frac{16}{25} = -\frac{36}{25}$$

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

$$u^2 = \frac{36}{25} u = \pm\sqrt{\frac{36}{25}} = \pm \frac{6}{5}$$

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

$$r =-\frac{4}{5} - \frac{6}{5} = -2 s = -\frac{4}{5} + \frac{6}{5} = 0.400$$

Evaluate

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