Factor the expression by grouping. First, the expression needs to be rewritten as $p^{2}+ap+bp+10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-7$$ $$ab=1\times 10=10$$
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 $10$.
$$-1,-10$$ $$-2,-5$$
Calculate the sum for each pair.
$$-1-10=-11$$ $$-2-5=-7$$
The solution is the pair that gives sum $-7$.
$$a=-5$$ $$b=-2$$
Rewrite $p^{2}-7p+10$ as $\left(p^{2}-5p\right)+\left(-2p+10\right)$.
$$\left(p^{2}-5p\right)+\left(-2p+10\right)$$
Factor out $p$ in the first and $-2$ in the second group.
$$p\left(p-5\right)-2\left(p-5\right)$$
Factor out common term $p-5$ by using distributive property.
$$\left(p-5\right)\left(p-2\right)$$
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 $ax^{2}+bx+c=0$.
$$p^{2}-7p+10=0$$
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.
Now solve the equation $p=\frac{7±3}{2}$ when $±$ is plus. Add $7$ to $3$.
$$p=\frac{10}{2}$$
Divide $10$ by $2$.
$$p=5$$
Now solve the equation $p=\frac{7±3}{2}$ when $±$ is minus. Subtract $3$ from $7$.
$$p=\frac{4}{2}$$
Divide $4$ by $2$.
$$p=2$$
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 $2$ for $x_{2}$.
$$p^{2}-7p+10=\left(p-5\right)\left(p-2\right)$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -7x +10 = 0$$
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 $rs = C$
$$r + s = 7 $$ $$ rs = 10$$
Two numbers $r$ and $s$ sum up to $7$ exactly when the average of the two numbers is $\frac{1}{2}*7 = \frac{7}{2}$. 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 $u$.
$$r = \frac{7}{2} - u$$ $$s = \frac{7}{2} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 10$
$$(\frac{7}{2} - u) (\frac{7}{2} + u) = 10$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{49}{4} - u^2 = 10$$
Simplify the expression by subtracting $\frac{49}{4}$ on both sides
$$-u^2 = 10-\frac{49}{4} = -\frac{9}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
