I can't understand how to factor a quadratic

Question

I can't figure out why some actions are possible in factoring $\frac{r–3}{r^2+11r–42}$. Factor the quadratic out of the denominator $\frac{r-3}{(r-3)(r+14)}$. This does not make sense how can $+11$ be removed and thrown into the trash?

Answer 1

The fact that it's in the denominator is not important. All this is really saying is that:

$$ r^2 + 11r - 42 = (r - 3)(r + 14) $$

We can verify this by hand by multiplying out the right-hand side: $$ \begin{align} (r - 3)(r + 14) &= r(r - 3) + 14(r - 3)\\ &= r^2 - 3r + 14r - 42 \\ &= r^2 + 11r - 42 \end{align} $$ as claimed.

How do you determine this? Well, there are a few ways. If you know the quadratic formula, this will tell you the roots of the polynomial---in our case, the numbers 3 and $-14$---which you can use to figure this out.

If you don't know the quadratic formula, and your teacher isn't too cruel, then the standard was is as follows. We want to write

$$ r^2 + 11r - 42 = (r - a)(r - b) $$

and we want to know what $a, b$ are. If we multiply out the right-hand side as above, we find that we have $$ (r - a)(r - b) = r^2 - (a + b)r + ab $$ and so if it is equal to $r^2 + 11r - 42$ then we must have that $$ ab = -42 \qquad \text{and} \qquad a + b = -11 $$ This is where it comes down to the cruelty of your teacher. Assuming they are kind and gentle souls, then we are hoping that $a, b$ will be integers and so we look at factors of -42. These are: $$ \begin{align} a = 1, b = -42 \\ a = 2, b = -21 \\ a = 3, b = -14 \\ a = 6, b = -7 \\ a = 7, b = -6\\ a = 14, b = -3 \\ a = 21, b = -2 \\ a = 42, b = -1 \end{align} $$ and then since we also want $a + b = -11$ we look through those pairs to find if any of them satisfy this. Lo and behold, $a = 3, b = -14$ work, and so we find our answer.

Answer 2

If we have a quadratic function $ax^2+bx+c$, we can rewrite it as $k(x-x_0)(x-x_1)$ for some constants $k$, $x_0$, $x_1$. We can do this because if we multiply this formula out, we obtain $$kx^2-kxx_0-kxx_1+kx_0x_1=kx^2-k(x_0+x_1)x+kx_0x_1.$$ Since we can choose any $k$, $x_0$, $x_1$, we can choose them such that $k=a$, $-k(x_0+x_1)=b$, and $kx_0x_1=c$. It is possible to extract from this an explicit formula, known as the quadratic formula: $$x_0=\frac{-b+\sqrt{b^2-4ac}}{2a},x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ but we may also factor many quadratic functions, particularly ones with $a=1$ such as in the given example, by the following method: if we have $x^2+bx+c$, find two numbers $z_0$ and $z_1$ such that $z_0+z_1=b$ and $z_0z_1=c$. Then $x^2+bx+c=(x+z_0)(x+z_1)$. (To see this, simply expand the product.)

In your case, we have $x^2+11x-42$, so we need two numbers which have sum $11$ and product $-42$. Since $14$ and $-3$ fulfill those requirements, $x^2+11x-42=(x-3)(x+11)$. The $11x$ does not disappear; it results from the sum of the products of $-3\cdot x$ and $14\cdot x$ when we expand the factored form.

EDIT: In response to the added question in the comments, this is called the distributive property of multiplication over addition: $a(b+c)=ab+ac$. It can be seen by imagining a rectangle with length $b+c$ and width $a$. The area of the rectangle is equal to $a(b+c)$, since the area of a rectangle is the product of its width and length. However, if we split the rectangle down its width so that we have two rectangles both with width $a$ but one having length $b$ and one having length $c$, then the sum of their areas is equal to $ab+ac$. Since the area of the whole rectangle is equal to the sum of the areas of the smaller rectangles, $a(b+c)$ must be equal to $ab+ac$.

Answer 3

I'll flesh out my comment a little.

The $+11$ term isn't thrown out as you say, the coefficient of $r$ depends on the sum of two terms when one factorises a quadratic. In general: \begin{eqnarray} (x+a)(x+b)&=&x^{2}+ax+bx+ab \\ &=& x^{2}+(a+b)x+ab \end{eqnarray} Wo when one factorises, we must find two numbers $a, b$ such that their sum equals the term in $x$ and their product equals the constant term $ab$.

Here, we must solve \begin{eqnarray} a+b &=& 11 \\ ab &=& -42 \end{eqnarray} $-42$ can be written as $\pm 3 \times \mp 14$, $\pm 6 \times \mp 7$, $\pm 1 \times \mp 42$ and $\pm 2 \times \mp 21$.

The only pair that will allow for a sum to $11$ is $-3+14$.