q(0)=p(2) ?
Let p(x) = x^4 + ax^3 + bx^2 + cx + d, where a, b, c, d are real numbers. It is known that...
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blamocur
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That's right. For which other values of [imath]x[/imath] we know [imath]q(x)[/imath] ?
I mean that my initial question (the one I posted) probably didn't end up with the answer he gave me , but the one dr.Peterson posted did produce the answer (5621) using my super long substitution steps.
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sir this is where I am lost, I can't form any connection here. T-T
do you mean I should make p(9) as p(3+6) since I have p(3)=523?
You are given three points that lie on the graph of p: (1,841), (2,1682), and (3, 523). Plugging those into the equation and solving for a, b, c, and d takes a lot of work (and big numbers with many places to make mistakes). But with experience, we know that it would be easier to solve a problem like this if the values of x were close to zero, and -1, 0, 1 would be really helpful. That suggests shifting the graph left by 2, so the points become (-1,841), (0,1682), and (1, 523). These would be points on the graph of a new function q(x) = p(x+2).
Have you learned about transforming graphs? This is a "shift" or "translation" by 2 units to the left.
Now try doing the same sort of thing you were doing, with these new points, to find an equation of q(x). Then, for the final answer, you'll need to find p(9) and p(-5). What values of x do those correspond to on the graph of q? You'll find something interesting, which makes the problem even easier.
@Dr.Peterson thank you for the examples
I am trying this q(-1) = 841 = 1 - a + b - c + d
q(0) = 1682 = d
q(1) = 523 = 1 + a + b + c + d
@blamocur is this also what you mean ?
Please tell me if I have the correct start...
Have you learned about transforming graphs? This is a "shift" or "translation" by 2 units to the left.
Now try doing the same sort of thing you were doing, with these new points, to find an equation of q(x). Then, for the final answer, you'll need to find p(9) and p(-5). What values of x do those correspond to on the graph of q? You'll find something interesting, which makes the problem even easier.
@Dr.Peterson thank you for the examples
I am trying this q(-1) = 841 = 1 - a + b - c + d
q(0) = 1682 = d
q(1) = 523 = 1 + a + b + c + d
@blamocur is this also what you mean ?
Please tell me if I have the correct start...
Dr.Peterson
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You have a system of two equations in three unknowns; that's not enough to fully solve for a, b, and c, but by subtracting the equations, you can eliminate a and c, and solve for b.
But what is p(9)? Isn't it q(9-2) = q(7) = 7^4 + 7^3a + 7^2b + 7c + d? Keep in mind that a, b, c, d are now the coefficients of q, not of p.
I haven't checked all the details yet, so there may be errors I haven't pointed out.
blamocur
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This is exactly what we meant.
But at this stage [imath]a,b,c,d[/imath] are coefficients for [imath]q[/imath], not [imath]p[/imath]. Instead of computing [imath]p(9) + p(-5)[/imath] you need to compute [imath]q(?)+q(??)[/imath].
@Dr.Peterson & @blamocur Thank you for helping me going through this. Appreciate it so much 
I am following up again from my workings above after reading your tips.
I solved for b = -1001
then set q(7) = 2401 + 343a + 49b + 7c + d
and also q(-7)= 2401 -343a + 49b -7c + d
I added both which gives 4802 + 98b + 2d
I had b = -1001 and d = 1682, giving me -89932
as for my initial post, I just need to minus that with 2 then divide by -8 giving 11241.75
as for dr.Peterson's post, I just need to minus that with 4 and divide by -16 giving 5621
Did I nail this?
I am following up again from my workings above after reading your tips.
I solved for b = -1001
then set q(7) = 2401 + 343a + 49b + 7c + d
and also q(-7)= 2401 -343a + 49b -7c + d
I added both which gives 4802 + 98b + 2d
I had b = -1001 and d = 1682, giving me -89932
as for my initial post, I just need to minus that with 2 then divide by -8 giving 11241.75
as for dr.Peterson's post, I just need to minus that with 4 and divide by -16 giving 5621
Did I nail this?
blamocur
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I believe you did -- congratulations.Did I nail this?
P.S. Not sure about the last line "as for dr.Peterson's post, I just need to minus that with 4 and divide by -16 giving 5621", but the 11241.75 matches what I got.
Dr.Peterson
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I got the same answers. I'll check by showing the work here.
Shifting the function to simplify the inputs, let [imath]q(x) = p(x+2)[/imath], and knowing that [imath]q(x)[/imath] will still be a monic quartic function, we have a new problem:
The conditions yield the equations [math](-1)^4+(-1)^3a+(-1)^2b+(-1)c+d=841\\(0)^4+(0)^3a+(0)^2b+(0)c+d=1682\\(1)^4+(1)^3a+(1)^2b+(1)c+d=523[/math] which simplify to [math]1-a+b-c+d=841\\d=1682\\1+a+b+c+d=523[/math] and then to [math]-a+b-c+d=840\\d=1682\\a+b+c+d=522[/math]
Adding the first and third equations, [math]2b+2d=1362\\b+d=681\\b+1682=681\\b=-1001[/math]
Therefore, [imath]q(x)=x^4+ax^3-1001x^2+cx+1682[/imath], so [math]q(7)=7^4+7^3a-1001\cdot7^2+7c+1682=343a+7c-44966\\q(-7)=7^4-7^3a-1001\cdot7^2-7c+1682=-343a-7c-44966[/math]
So the answers to the problems are
Yes, we agree.
Shifting the function to simplify the inputs, let [imath]q(x) = p(x+2)[/imath], and knowing that [imath]q(x)[/imath] will still be a monic quartic function, we have a new problem:
Given that [imath]q(x)=x^4+ax^3+bx^2+cx+d[/imath], and that [imath]q(-1)=841,q(0)=1682,\text{ and }q(1)=523[/imath], find
(a) [imath]\dfrac{q(7)+q(-7)-2}{-8}[/imath], and
(b) [imath]\dfrac{q(7)+q(-7)-4}{-16}[/imath].
The conditions yield the equations [math](-1)^4+(-1)^3a+(-1)^2b+(-1)c+d=841\\(0)^4+(0)^3a+(0)^2b+(0)c+d=1682\\(1)^4+(1)^3a+(1)^2b+(1)c+d=523[/math] which simplify to [math]1-a+b-c+d=841\\d=1682\\1+a+b+c+d=523[/math] and then to [math]-a+b-c+d=840\\d=1682\\a+b+c+d=522[/math]
Adding the first and third equations, [math]2b+2d=1362\\b+d=681\\b+1682=681\\b=-1001[/math]
Therefore, [imath]q(x)=x^4+ax^3-1001x^2+cx+1682[/imath], so [math]q(7)=7^4+7^3a-1001\cdot7^2+7c+1682=343a+7c-44966\\q(-7)=7^4-7^3a-1001\cdot7^2-7c+1682=-343a-7c-44966[/math]
So the answers to the problems are
(a) [imath]\dfrac{(343a+7c-44966)+(-343a-7c-44966)-2}{-8}=\dfrac{2(-44966)-2}{-8}=11241.75[/imath], and
(b) [imath]\dfrac{(343a+7c-44966)+(-343a-7c-44966)-4}{-16}=\dfrac{2(-44966)-4}{-16}=5621[/imath].
Yes, we agree.
blamocur
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Never mind, I haven't noticed slightly different formula in post #5.
Thank you so much @blamocur and @Dr.Peterson for your advice on the "short/quick/better" method. I used the long algebraic substitution method at first and yes it took a very long time and prone to mistakes. I am reflecting on the process, even though I learned horizontal translation of functions in class but this is my first time realising it can be applied in such situation to solve & simplify algebraic related problems.
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