Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-40$. 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 negative, the negative number has greater absolute value than the positive. 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 $-3$.
$$a=-8$$ $$b=5$$
Rewrite $x^{2}-3x-40$ as $\left(x^{2}-8x\right)+\left(5x-40\right)$.
$$\left(x^{2}-8x\right)+\left(5x-40\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-8\right)+5\left(x-8\right)$$
Factor out common term $x-8$ by using distributive property.
$$\left(x-8\right)\left(x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}-3x-40=0$$
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.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$x^{2}-3x-40=\left(x-8\right)\left(x+5\right)$$
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 $u$.
$$r = \frac{3}{2} - u$$ $$s = \frac{3}{2} + u$$
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 $u$
