Factoring the polynomial $5c^2 - 52c + 20$

Question

I was trying to find the factors of that expression above:

$5c^2-52c+20.$

The solution is

$(5c−2)(c−10).$

I don't understand: how did we get these values $-2$ and $-10?$ Is there any way to solve it quickly by looking at some parts of that expression?

Answer 1

$$5c^2-52c+20=0$$ $$5c^2-50c-2c+20=0\\5c(c-10)-2(c-10)=0$$ $$(5c-2)(c-10)=0$$ or you can use quadratic formula $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$ Here $a=5,b=-52,c=20$

Answer 2

Break the expression into the groups

$$\left(5c^2-50c\right)+\left(-2c+20\right)$$

Factor out $5c$ from $5c^2-50$, which yields $5c\left(c-10\right)$. Now factor out $-2$ from $-2c+20$ and we get $-2\left(c-10\right)$

Factor out the common term $\left(c-10\right)$

$$\left(c-10\right)\left(5c-2\right)$$

Answer 3

There's a generic algorithm that works for these types of trinomials ($ax^2+bx+c$):

1. First write $\left(\dfrac{ax\phantom{+4}}{\phantom{5}}\right) \left(\dfrac{ax\phantom{+4}}{\phantom{5}}\right).$
2. Find the product $ac$, including sign.
3. Find the prime factorization of $ac$ using the factor tree.
4. Find all factor pairs of $ac$ using the factor tree: begin with $1,ac$, and increase the $1$ according to whether you can get it by a product of numbers in the prime factorization of $ac$. You are done when the first number has reached $\sqrt{ac}$.
5. Find the factor pair of $ac$ such that the two numbers add to $b$, including sign. If $ac>0$, then the two numbers will have the same sign. If $ac<0$, then the two numbers will have opposite signs. (If this step fails, the quadratic does not factor.) Call these two numbers $s$ and $t$.
6. Write $\left(\dfrac{ax+s}{\phantom{5}}\right)\left(\dfrac{ax+t}{\phantom{5}}\right).$
7. Divide each of these binomials by its own GCF: $\left(\dfrac{ax+s}{\operatorname{gcf}(a,s)}\right) \left(\dfrac{ax+t}{\operatorname{gcf}(a,t)}\right).$ Check that $$\operatorname{gcf}(a,s)\cdot\operatorname{gcf}(a,t)=a.$$

Answer 4

Consider what happens when you multiply two linear polynomials.

\begin{align*} (4x - 5)(3x + 8) & = 4x(3x + 8) - 5(3x + 8) && \text{apply the distributive law}\\ & = \color{blue}{12}x^2 + \color{green}{32}x - \color{green}{15}x - \color{blue}{40} && \text{apply the distributive law}\\ & = \color{blue}{12}x^2 + \color{green}{17}x - \color{blue}{40} && \text{combine like terms} \end{align*} Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients that combine to form the linear coefficient of the product, that is, $$(\color{blue}{12})(\color{blue}{-40}) = (\color{green}{32})( \color{green}{-15})$$ To factor $12x^2 + 17x - 40$, we wish to carry out these steps in reverse.
\begin{align*} 12x^2 + \color{green}{17}x - 40 & = 12x^2 + 32x - 15x - 40 && \text{split the linear term}\\ & = 4x(3x + 8) - 5(3x + 8) && \text{factor by grouping}\\ & = (4x - 5)(3x + 8) && \text{extract the common factor} \end{align*} The observation we made above enables us to split the linear term. To do so, we must find two numbers with product $(\color{blue}{12})(\color{blue}{-40}) = -480$ that have sum $\color{green}{17}$. We can do so by listing the factor pairs of $-480$, then finding the pair in which the sum of the factors is $\color{green}{17}$.

\begin{array}{c c c c} \text{factorization} & \text{factor sum} & \text{factorization} & \text{factor sum}\\ \hline (1)(-480) & -479 & (-1)(480) & 479\\ (2)(-240) & -238 & (-2)(240) & 238\\ (3)(-160) & -157 & (-3)(160) & 157\\ (4)(-120) & -116 & (-4)(120) & 116\\ (5)(-96) & -91 & (-5)(96) & 91\\ (6)(-80) & -74 & (-6)(80) & 74\\ (8)(-60) & -52 & (-8)(60) & 52\\ (10)(-48) & -38 & (-10)(48) & 38\\ (12)(-40) & -28 & (-12)(40) & 28\\ (15)(-32) & -17 & (\color{green}{-15})(\color{green}{32}) & \color{green}{17}\\ (20)(-24) & -4 & (-20)(24) & 4 \end{array}
The desired factors are $\color{green}{32}$ and $\color{green}{-15}$. Once we split the linear term, we extract the common factor from the first two terms and the common factor from the last two terms, which is called factoring by grouping. Finally, we extract the common linear factor to complete the factorization.

Let's apply this technique to your example.

$$5c^2 - 52c + 20$$

To split the linear term, we must find two numbers with product $\color{blue}{5} \cdot \color{blue}{20} = 100$ and sum $\color{green}{-52}$. We list the factor pairs of $100$.

\begin{array}{c c c c} \text{factorization} & \text{factor sum} & \text{factorization} & \text{factor sum}\\ \hline 1 \cdot 100 & 101 & (-1)(-100) & -101\\ 2 \cdot 50 & 52 & (\color{green}{-2})(\color{green}{-50}) & \color{green}{-52}\\ 4 \cdot 25 & 29 & (-4)(-25) & -29\\ 5 \cdot 20 & 25 & (-5)(-20) & -25\\ 10 \cdot 10 & 20 & (-10)(-10) & -20 \end{array} The desired factors are $\color{green}{-2}$ and $\color{green}{-50}$. Hence, \begin{align*} 5c^2 - 52c + 20 & = 5c^2 - 2c - 50c + 20 && \text{split the linear term}\\ & = c(5c - 2) - 10(5c - 2) && \text{factor by grouping}\\ & = (c - 10)(5c - 2) && \text{extract the common factor} \end{align*}

More generally, a factorization $$ax^2 + bx + c = (rx + s)(tx + u)$$ of a quadratic polynomial $ax^2 + bx + c$ with respect to the rational numbers can be found if $a$, $b$, and $c$ are integers such that there exist integers with product $ac$ and sum $b$.

Caveat: Almost every quadratic polynomial you can write down is irreducible with respect to the rational numbers, meaning it cannot be split into two linear factors with rational coefficients. For instance, $x^2 + 2x + 3$ is irreducible with respect to the rational numbers since there are no integers with product $1 \cdot 3 = 3$ and sum $2$.

Answer 5

There's always completing the square and/or the quadratic formula.

And bear in mind not all quadratic expression factor to rational components.

But:

In general:

if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = \frac {-b +\sqrt{b^2 -4ac}}{2a}$ or $x = k_2 = \frac {-b - \sqrt{b^2 - 4ac}}{2a}$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.

Let's explain:

So if we solve $5c^2 - 52c + 20 = 0$ by completing the square (or quadratic formula) we get:

$5c^2 - 52c + 20 = 0$

$c^2 - \frac {52}{5}c + 4 = 0$

$c^2 - \frac {52}{5}c = -4$

$c^2 - \frac {52}{5}c + (\frac {26}5)^2 $

$(c - \frac {26}5)^2 = = (\frac {26}5)^2 - 4=\frac {4*13^2}{25} - \frac {4*25}{25}= \frac {4(13^2 - 5^2)}{25} = \frac {4*144}{25}=(\frac {2*12}5)^2$

$c -\frac {26}5 = \pm \frac {24}{5}$

$c = \frac {26}5 \pm {24}5$

$c = \frac 25$ or $c = 10$ so

.....

Now

Suppose $5c^2 - 52c + 20$ factored into $(mc + u)(nc + v)=mn(c +\frac um)(c+\frac vn)$. Then $5c^2 - 52c + 20 = 0 \iff (c +\frac um)(c+\frac vn) = 0 \iff c= -\frac um$ or $c = -\frac vn$.

So we know that the solutions are $c = \frac 25$ and $c = 10$ so we must have that $5c^2 - 52c + 20$ factors to $5(c-\frac 25)(c-10)$ or $(5c - 2)(c-20)$.

In general:

if the solution to $ax^2 + bx + c = 0$ is $x= k_1 = \frac {-b +\sqrt{b^2 -4ac}}{2a}$ or $x = k_2 = \frac {-b - \sqrt{b^2 - 4ac}}{2a}$, then $ax^2 + bx + c$ will factor to $a(x-k_1)(x-k_2)$.

....

That's how you do it if you just can't see anything else.

In general you can make educated guesses:

If $5c^2 + 52c + 20 = (mc + u)(nc+ v)$ then

$mn = 5$

$un + vm = -52$

$uv = 20$.

If this behaves "nicely" (which you have no reason to assume it will!) then $mn = 5$ means $m,n = 5,1$.

And $uv = 20$ so $u,v =\pm 1,20, \pm 2,10, \pm 4,5$

We have $u + 5v = -52$. The mean $u = -2, - 7 ,-12....$ and $v = -10, -9, -8....$ etc. The only choses we have that work are $-2,-10$.

So it was a very educated guess that $(5c -2)(c-10) =5c^2 + (-2-5*10)c +(-2*-10) = 5c^2 -52c + 20$.

Answer 6

$$\begin{align} 5c^2 - 52c + 20 &= \frac{5}{5}\cdot\big(5c^2 - 52c + 20\big)\\ &= \frac{25c^2 - 260c + 100}{5}\\ &= \frac{(5c)^2 -52(5c) + 100}{5}\\ &= \frac{t^2 -52t + 100}{5}\text{ where } t=5c\\ \end{align}$$

Can you take it from there?