Factor the expression by grouping. First, the expression needs to be rewritten as $q^{2}+aq+bq+21$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-10$$ $$ab=1\times 21=21$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $21$.
$$-1,-21$$ $$-3,-7$$
Calculate the sum for each pair.
$$-1-21=-22$$ $$-3-7=-10$$
The solution is the pair that gives sum $-10$.
$$a=-7$$ $$b=-3$$
Rewrite $q^{2}-10q+21$ as $\left(q^{2}-7q\right)+\left(-3q+21\right)$.
$$\left(q^{2}-7q\right)+\left(-3q+21\right)$$
Factor out $q$ in the first and $-3$ in the second group.
$$q\left(q-7\right)-3\left(q-7\right)$$
Factor out common term $q-7$ by using distributive property.
$$\left(q-7\right)\left(q-3\right)$$
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 $ax^{2}+bx+c=0$.
$$q^{2}-10q+21=0$$
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.
Now solve the equation $q=\frac{10±4}{2}$ when $±$ is plus. Add $10$ to $4$.
$$q=\frac{14}{2}$$
Divide $14$ by $2$.
$$q=7$$
Now solve the equation $q=\frac{10±4}{2}$ when $±$ is minus. Subtract $4$ from $10$.
$$q=\frac{6}{2}$$
Divide $6$ by $2$.
$$q=3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $7$ for $x_{1}$ and $3$ for $x_{2}$.
$$q^{2}-10q+21=\left(q-7\right)\left(q-3\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -10x +21 = 0$$
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 $rs = C$
$$r + s = 10 $$ $$ rs = 21$$
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 $u$.
$$r = 5 - u$$ $$s = 5 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 21$
$$(5 - u) (5 + u) = 21$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$25 - u^2 = 21$$
Simplify the expression by subtracting $25$ on both sides
$$-u^2 = 21-25 = -4$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 4$$ $$u = \pm\sqrt{4} = \pm 2 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
