Factor the expression by grouping. First, the expression needs to be rewritten as $n^{2}+an+bn-10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\left(-10\right)=-10$$
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 $-10$.
$$-1,10$$ $$-2,5$$
Calculate the sum for each pair.
$$-1+10=9$$ $$-2+5=3$$
The solution is the pair that gives sum $3$.
$$a=-2$$ $$b=5$$
Rewrite $n^{2}+3n-10$ as $\left(n^{2}-2n\right)+\left(5n-10\right)$.
$$\left(n^{2}-2n\right)+\left(5n-10\right)$$
Factor out $n$ in the first and $5$ in the second group.
$$n\left(n-2\right)+5\left(n-2\right)$$
Factor out common term $n-2$ by using distributive property.
$$\left(n-2\right)\left(n+5\right)$$
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 $ax^{2}+bx+c=0$.
$$n^{2}+3n-10=0$$
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.
$$n=\frac{-3±\sqrt{3^{2}-4\left(-10\right)}}{2}$$
Square $3$.
$$n=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$
Multiply $-4$ times $-10$.
$$n=\frac{-3±\sqrt{9+40}}{2}$$
Add $9$ to $40$.
$$n=\frac{-3±\sqrt{49}}{2}$$
Take the square root of $49$.
$$n=\frac{-3±7}{2}$$
Now solve the equation $n=\frac{-3±7}{2}$ when $±$ is plus. Add $-3$ to $7$.
$$n=\frac{4}{2}$$
Divide $4$ by $2$.
$$n=2$$
Now solve the equation $n=\frac{-3±7}{2}$ when $±$ is minus. Subtract $7$ from $-3$.
$$n=-\frac{10}{2}$$
Divide $-10$ by $2$.
$$n=-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $2$ for $x_{1}$ and $-5$ for $x_{2}$.
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
$$n^{2}+3n-10=\left(n-2\right)\left(n+5\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 +3x -10 = 0$$
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 $rs = C$
$$r + s = -3 $$ $$ rs = -10$$
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 $rs = -10$
$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -10$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{9}{4} - u^2 = -10$$
Simplify the expression by subtracting $\frac{9}{4}$ on both sides
$$-u^2 = -10-\frac{9}{4} = -\frac{49}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
