Solve q^2+3q-40

$$q ^ { 2 } + 3 q - 40$$

Factor

$$\left(q-5\right)\left(q+8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as q^{2}+aq+bq-40. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-40\right)=-40$$

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 -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 3.

$$a=-5 b=8$$

Rewrite q^{2}+3q-40 as \left(q^{2}-5q\right)+\left(8q-40\right).

$$\left(q^{2}-5q\right)+\left(8q-40\right)$$

Factor out q in the first and 8 in the second group.

$$q\left(q-5\right)+8\left(q-5\right)$$

Factor out common term q-5 by using distributive property.

$$\left(q-5\right)\left(q+8\right)$$

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.

$$q^{2}+3q-40=0$$

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.

$$q=\frac{-3±\sqrt{3^{2}-4\left(-40\right)}}{2}$$

Square 3.

$$q=\frac{-3±\sqrt{9-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$q=\frac{-3±\sqrt{9+160}}{2}$$

Add 9 to 160.

$$q=\frac{-3±\sqrt{169}}{2}$$

Take the square root of 169.

$$q=\frac{-3±13}{2}$$

Now solve the equation q=\frac{-3±13}{2} when ± is plus. Add -3 to 13.

$$q=\frac{10}{2}$$

Divide 10 by 2.

$$q=5$$

Now solve the equation q=\frac{-3±13}{2} when ± is minus. Subtract 13 from -3.

$$q=-\frac{16}{2}$$

Divide -16 by 2.

$$q=-8$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -8 for x_{2}.

$$q^{2}+3q-40=\left(q-5\right)\left(q-\left(-8\right)\right)$$

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

$$q^{2}+3q-40=\left(q-5\right)\left(q+8\right)$$

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.

$$x ^ 2 +3x -40 = 0$$

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

$$r + s = -3 rs = -40$$

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -40

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -40$$

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

$$\frac{9}{4} - u^2 = -40$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -40-\frac{9}{4} = -\frac{169}{4}$$

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.

$$r =-\frac{3}{2} - \frac{13}{2} = -8 s = -\frac{3}{2} + \frac{13}{2} = 5$$

Evaluate

$$\left(q-5\right)\left(q+8\right)$$