# I need to know how to write a cubic function with a remainder of 8 for f(2) and a remainder of -5 for f(3) using synthetic division.

### 2 Answers | Add Yours

We are asked to find a cubic function such that f(2) has remainder 8 and f(3) has remainder -5.

Let the cubic be x^3+ax^2+bx+c. (There are an infinite number of cubics through 2 points -- we will choose one with leading coefficient 1.)

Using synthetic division:

2 | 1 a b c

------------------------------

1 (a+2) 2a+b+4 4a+2b+c+8

So we know 4a+2b+c=0. Now try 3:

3 | 1 a b c

------------------------------------

1 a+3 3a+b+9 9a+3b+c+27

Then 9a+3b+c=-32

We have 2 equations in 3 unknowns -- solving the system we find that there are an infinite number of possible solutions (as expected) of the form:

(a,-5a-32,6a+64)

Let a=1; then b=-37 and c=70 to get a cubic:

`f(x)=x^3+x^2-37x+70 `

Note that f(2)=8 and f(3)=-5 as required.

Let a=3; then b=-47 and c=82.

This cubic is `f(x)=x^3+3x^2-47x+82 `

---------------------------------------------------------------------------

One solution to the problem is `f(x)=x^3+x^2-37x+70 `

We can characterize all solutions with leading coefficient 1:

`f(x)=x^3+ax^2+bx+c ` where a is any real number, b=-5a-32 and c=6a+64. There are other answers where the leading coefficient is not 1.

---------------------------------------------------------------------------

**Sources:**

Work your way backwards.

For a cubic function with a remainder of 8 for f(2), we know that:

2) A B C D

+ 2A 2E 2F

-------------------

A E F 8

and that Ax^3+Bx^2+Cx+D will be the equation.

Now plug in any number in F, and determine what D would be with D+2F being 8.

For example, if you plug 10 into F, it would be:

2) A B C -12

+ 2A 2E 20

-------------------

A E 10 8

now plug another number into E, for instance 9:

2) A B -8 -12

+ 2A 18 20

-------------------

A 9 10 8

plug another number into A, for instance 4:

2) 1 1 -8 -12

+ 8 18 20

-------------------

1 9 10 8

Now, the equation would be:

x^3+x^2-8x-12

Hope this helped, now you can try the latter yourself!

` `

### Hide Replies ▲

Your set up is correct, but in synthetic division you add, not subtract. You can check your answer to see that it does not meet the given conditions.