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Given the following: $$y = a\,\left( - c+x\right) \,\left( - d+x\right)$$ and that "the co-efficient of $x$ is $b$", and asked to define $a$ in terms of $c$ and $d$, why is the answer: $$a = \frac{-b}{c+d}$$

When I attempt to solve this myself, I end up with a series of unlike terms. What happened to $x$? Is the book I’m working from wrong?

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    $\begingroup$ I think I understand but I would like to clarify something before I make my assumption. Do you mean to have a negative anywhere in that fraction? $\endgroup$
    – randomgirl
    Nov 19, 2022 at 4:59
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    $\begingroup$ I will go ahead and make my assumption. If you write your polynomial in standard form (expanded form), you should see the coefficient is something else other than b. So you will set that expression equal to b and solve for a. $\endgroup$
    – randomgirl
    Nov 19, 2022 at 5:02
  • $\begingroup$ Nothing happened to $x$. If you expan $a(-c+x)(-d+x)$ you are going to get something of the form $y = \alpha x^2 + \beta x + \gamma$ where $\alpha$, $\beta$ and $\gamma$ are all some combination of $a,d,c$. The coefficient of $x$, what I labeled as $\beta$, is some combination of $a,c,d$. The book is calling that coefficient $b$. So you should have $b = $some combination of $a,c,d$. Solve for $a$. $\endgroup$
    – fleablood
    Nov 19, 2022 at 5:34
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    $\begingroup$ "when I attempt to solve this myself i end up with a series of unlike terms" what do you end up with. We can't tell you what you did wrong (and yes, you, did something wrong, not the book) if you don't tell us what you did. $\endgroup$
    – fleablood
    Nov 19, 2022 at 5:35

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Have $y = a(x-c)(x-d)$
$$\implies y = a(x^2 - (c+d)x +cd) = ax^2 - a(c+d)x +acd)$$ We are told the coefficient of $x$ is $b$ so we have: $$-a(c+d) = b$$ and then rearranging for $a$ gives: $$a=\frac{-b}{c+d}$$

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  • $\begingroup$ Thanks @Ogglie! I understand your expansion, but there is something im missing which makes it difficult for me to follow the part about the co-efficient of x being b. If the co-efficient of x is b, then isnt the next step to substitute $-a(c+d)$ with $b$? Can you elaborate on the rearrangement process? $\endgroup$
    – duckegg
    Nov 19, 2022 at 21:42
  • $\begingroup$ The penny just dropped. Given the answer, the instruction to define a in terms of c and d appears incomplete, or misleading. Perhaps i am still too unfamiliar with the conventions of mathematical discourse to have understood that $x$ and $y$ could be ignored, also the rightmost term in the expansion $cd$. I see how it is possible to equate b with the middle term and solve for $a$, but I'm confused about where in the challenge it is permissible to ignore all other terms, or is this implicit in the operation? $\endgroup$
    – duckegg
    Nov 19, 2022 at 22:38
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    $\begingroup$ I think I understand your confusion. Let's say we have that: $$ax^2 + bx +c = px^2 + qx +r$$ It's important to understand that $x$ is a variable so this relation holds true for any value of $x$. Rearranging gives: $$(a-p)x^2+(b-q)x+c-r = 0$$ Logically we can see that for this to be true no matter what the value of $x$ is, we must have $a=p,b=q,c=r$ but I will prove this regardless. This is true for all x so let's set $x=0$, then: $$c-r=0 \implies c=r$$ Substituting this back tells us: $$(a-p)x^2 + (b-q)x = 0$$ $\endgroup$ Nov 20, 2022 at 15:17
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    $\begingroup$ Now let's set $x=1$: $$a-p+b-q=0 \implies a-p = -(b-q)$$ Again, we substitute this back into our quadratic: $$-(b-q)x^2 + (b-q)x = 0$$ If we now set $x=2$: $$-4(b-q) + (b-q) = 0$$ rearranging gives: $$b=q$$ So, we can conclude that if two quadratics are equal, so are their $x$ coefficients, and, in fact, their $x^2$ and $x^0$ coefficients (by the same reasoning). Therefore, if the $x$ coefficient of $y$ is $b$, since $y = ax^2 - a(c+d)x +acd$, we must have: $b = -a(c+d)$ $\endgroup$ Nov 20, 2022 at 15:17
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    $\begingroup$ I'm not exactly sure what you mean by this. If you mean noticing that b=q then this is called "comparing coefficients". $\endgroup$ Nov 21, 2022 at 3:25

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