Solve y^2-5y-24

$$y ^ { 2 } - 5 y - 24$$

Factor

$$\left(y-8\right)\left(y+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-24. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-24\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=-8 b=3$$

Rewrite y^{2}-5y-24 as \left(y^{2}-8y\right)+\left(3y-24\right).

$$\left(y^{2}-8y\right)+\left(3y-24\right)$$

Factor out y in the first and 3 in the second group.

$$y\left(y-8\right)+3\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(y+3\right)$$

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.

$$y^{2}-5y-24=0$$

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.

$$y=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-24\right)}}{2}$$

Square -5.

$$y=\frac{-\left(-5\right)±\sqrt{25-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$y=\frac{-\left(-5\right)±\sqrt{25+96}}{2}$$

Add 25 to 96.

$$y=\frac{-\left(-5\right)±\sqrt{121}}{2}$$

Take the square root of 121.

$$y=\frac{-\left(-5\right)±11}{2}$$

The opposite of -5 is 5.

$$y=\frac{5±11}{2}$$

Now solve the equation y=\frac{5±11}{2} when ± is plus. Add 5 to 11.

$$y=\frac{16}{2}$$

Divide 16 by 2.

$$y=8$$

Now solve the equation y=\frac{5±11}{2} when ± is minus. Subtract 11 from 5.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and -3 for x_{2}.

$$y^{2}-5y-24=\left(y-8\right)\left(y-\left(-3\right)\right)$$

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

$$y^{2}-5y-24=\left(y-8\right)\left(y+3\right)$$

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.

$$x ^ 2 -5x -24 = 0$$

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

$$r + s = 5 rs = -24$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -24

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -24$$

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

$$\frac{25}{4} - u^2 = -24$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -24-\frac{25}{4} = -\frac{121}{4}$$

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

$$u^2 = \frac{121}{4} u = \pm\sqrt{\frac{121}{4}} = \pm \frac{11}{2}$$

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

$$r =\frac{5}{2} - \frac{11}{2} = -3 s = \frac{5}{2} + \frac{11}{2} = 8$$

Evaluate

$$\left(y-8\right)\left(y+3\right)$$