Solve y^2+5y-36

$$y ^ { 2 } + 5 y - 36$$

Factor

$$\left(y-4\right)\left(y+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-36. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-36\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite y^{2}+5y-36 as \left(y^{2}-4y\right)+\left(9y-36\right).

$$\left(y^{2}-4y\right)+\left(9y-36\right)$$

Factor out y in the first and 9 in the second group.

$$y\left(y-4\right)+9\left(y-4\right)$$

Factor out common term y-4 by using distributive property.

$$\left(y-4\right)\left(y+9\right)$$

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.

$$y^{2}+5y-36=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-36\right)}}{2}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-36\right)}}{2}$$

Multiply -4 times -36.

$$y=\frac{-5±\sqrt{25+144}}{2}$$

Add 25 to 144.

$$y=\frac{-5±\sqrt{169}}{2}$$

Take the square root of 169.

$$y=\frac{-5±13}{2}$$

Now solve the equation y=\frac{-5±13}{2} when ± is plus. Add -5 to 13.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{-5±13}{2} when ± is minus. Subtract 13 from -5.

$$y=-\frac{18}{2}$$

Divide -18 by 2.

$$y=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and -9 for x_{2}.

$$y^{2}+5y-36=\left(y-4\right)\left(y-\left(-9\right)\right)$$

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

$$y^{2}+5y-36=\left(y-4\right)\left(y+9\right)$$

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.

$$x ^ 2 +5x -36 = 0$$

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

$$r + s = -5 rs = -36$$

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 = -36

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -36$$

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

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

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

$$-u^2 = -36-\frac{25}{4} = -\frac{169}{4}$$

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

$$u^2 = \frac{169}{4} u = \pm\sqrt{\frac{169}{4}} = \pm \frac{13}{2}$$

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

$$r =-\frac{5}{2} - \frac{13}{2} = -9 s = -\frac{5}{2} + \frac{13}{2} = 4$$

Evaluate

$$\left(y-4\right)\left(y+9\right)$$