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enter image description here Function is $f(x)= ax^4-25x^3+2x^2+25x+b $ it passes through (7,E) E= 1296 and there is a zero of -1

CiaPan
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    Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Brian Mar 22 '19 at 14:32
  • Welcome to stackexchange. When you [edit] the question to ask it directly rather than with a hard to read image, be sure to tell us what you tried and where you are stuck. Please use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Mar 22 '19 at 14:35
  • @EnriqueRodriguez Please add the function and your work into your question. If you show your effort appropriately, it is more likely that somebody will also put his/her effort into providing you some help. – Ertxiem - reinstate Monica Mar 22 '19 at 14:42
  • @Ertxiem is this better? – Enrique Rodriguez Mar 22 '19 at 15:22
  • @Brian i added the function is this better? – Enrique Rodriguez Mar 22 '19 at 15:23

1 Answers1

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You can find $a$ and $b$ as a system of equations:

We know that for $x=7$, $f(x)=1296$, this will be one equation on $a$ and $b$.

And for $x=-1$, $f(x) = 0$, this will give another equation on $a$ and $b$.

Can you write down the equations and solve the system or do you need more help?

Edit: The equations will be: $$\begin{cases} a \times 7^4 - 25 \times 7^3 + 2 \times 7^2+25 \times 7 + b = 1296 \\ a \times (-1)^4 - 25 \times (-1)^3 + 2 \times (-1)^2+25 \times (-1) + b = 0 \end{cases}$$

Edit 2: If you subtract the first equation from the second you get an equation without $b$: $$a \times (7^4-(-1)^4) - 25 \times (7^3-(-1)^3) + 2 \times (7^2-(-1)^2) + 25 \times (7-(-1)) = 1296 \ .$$ I was a bit lazy and I did not compute the powers of $7$. :)

With this last equation, you'll be able to compute the value of $a$. To obtain the value of $b$ you can use one of the other equations (I suggest using the one with the smaller numbers).