Solve y^2+10y-39

$$y ^ { 2 } + 10 y - 39$$

Factor

$$\left(y-3\right)\left(y+13\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by-39. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\left(-39\right)=-39$$

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 -39.

$$-1,39 -3,13$$

Calculate the sum for each pair.

$$-1+39=38 -3+13=10$$

The solution is the pair that gives sum 10.

$$a=-3 b=13$$

Rewrite y^{2}+10y-39 as \left(y^{2}-3y\right)+\left(13y-39\right).

$$\left(y^{2}-3y\right)+\left(13y-39\right)$$

Factor out y in the first and 13 in the second group.

$$y\left(y-3\right)+13\left(y-3\right)$$

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

$$\left(y-3\right)\left(y+13\right)$$

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.

$$y^{2}+10y-39=0$$

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.

$$y=\frac{-10±\sqrt{10^{2}-4\left(-39\right)}}{2}$$

Square 10.

$$y=\frac{-10±\sqrt{100-4\left(-39\right)}}{2}$$

Multiply -4 times -39.

$$y=\frac{-10±\sqrt{100+156}}{2}$$

Add 100 to 156.

$$y=\frac{-10±\sqrt{256}}{2}$$

Take the square root of 256.

$$y=\frac{-10±16}{2}$$

Now solve the equation y=\frac{-10±16}{2} when ± is plus. Add -10 to 16.

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

Divide 6 by 2.

$$y=3$$

Now solve the equation y=\frac{-10±16}{2} when ± is minus. Subtract 16 from -10.

$$y=-\frac{26}{2}$$

Divide -26 by 2.

$$y=-13$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -13 for x_{2}.

$$y^{2}+10y-39=\left(y-3\right)\left(y-\left(-13\right)\right)$$

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

$$y^{2}+10y-39=\left(y-3\right)\left(y+13\right)$$

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.

$$x ^ 2 +10x -39 = 0$$

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

$$r + s = -10 rs = -39$$

Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -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>

$$r = -5 - u s = -5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -39

$$(-5 - u) (-5 + u) = -39$$

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

$$25 - u^2 = -39$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -39-25 = -64$$

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

$$u^2 = 64 u = \pm\sqrt{64} = \pm 8$$

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

$$r =-5 - 8 = -13 s = -5 + 8 = 3$$

Evaluate

$$\left(y-3\right)\left(y+13\right)$$