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I was doing my homework and I am now stuck on question number 7 which is:

The diagram shows the curve with the equation $y = (x + a)(x - b)^2$ where $a$ and $b$ are positive integers.

(i) Write down the values of $a$ and $b$, and also of $c$, given that the curve crosses the $y$-axis at $(0, c)$.

I have attached a picture of my text book here:

enter image description here

I have so far found $b$ like this:

$(x - b)^2 = x^2 - 2bx + b^2$

Since $y = 0$ at $x = 1$,

$0 = 1 - 2b + b^2$

Solving that quadratic equation gives us $b = 1$.

I need to find $a$ and $c$ and I am totally puzzled on how to find them.

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  • $\begingroup$ For $a$ the solution is the same... only you have a first order equation. Then, to solve for $c$, put $x=0$ in the original equation. $\endgroup$ – N74 May 5 '16 at 14:20
  • $\begingroup$ @N74 My textbook says a and c are 2 $\endgroup$ – Hassan Althaf May 5 '16 at 14:21
  • $\begingroup$ You are told the equation is $(x+a)(x-b)^2$ with $a,b$ positive integers. Looking at the graph, the positive root must be at 1, so $b=1$. The negative root must be at $-2$, so $a=2$. Then the equation has constant term $ab^2=2$, so $c=2$. $\endgroup$ – almagest May 5 '16 at 14:29
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From the graph it is evident that the roots are $-2$ and $1$ ($1$ being a repeated root),

So the equation becomes: $$(x+2)(x-1)^2=0$$ Comparing it with $$(x+a)(x-b)^2=0$$ We find $a=2,b=1$

Also for finding the point $(0,c)$ we can plug in $x=0$ in the equation $y=(x+2)(x-1)^2$ and we get $c=2$

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  • $\begingroup$ Woah. Is it possible to find if for exanple we did not know it cuts - 2 $\endgroup$ – Hassan Althaf May 5 '16 at 14:29
  • $\begingroup$ @HassanAlthaf I don't get what you mean? $\endgroup$ – Nikunj May 5 '16 at 14:30
  • $\begingroup$ we know the curve cuts the x axis at -2, can we still find $a$ without that information? $\endgroup$ – Hassan Althaf May 5 '16 at 14:31
  • $\begingroup$ No, if you don't use another point of intersection, you cannot determine $a$, $\endgroup$ – Nikunj May 5 '16 at 14:33
  • $\begingroup$ I meant that if it still did intersect but we did not know that the x coordinate was '-2' $\endgroup$ – Hassan Althaf May 5 '16 at 14:34

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