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# Polynomial Divison

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Find a polynomial f(x) of degree 5 such that both of these properties hold:

f(x) is divisible by $$x^3$$.

f(x) +2 is divisible by $$(x+1)^3$$.

I've tried setting an equation of ax^5+bx^4+cx^3 because it can't have a x^2 term or lower to be divisible and doing some substitution..... but it doesn't work :P
Any help would be much appreciated!

May 7, 2022

#1
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Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

May 7, 2022
#2
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I don't think that's right

Guest May 7, 2022
#3
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I do not know if guests answer is right or wrong, I have not checked.

Let

$$f(x)=x^3(ax^2+bx+c)\quad so\\ f(x)+2=x^3(ax^2+bx+c)+2\qquad (1)\\ f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But also}\\~\\ f(x)+2=(x+1)^3(dx^2+ex+f)\qquad (2)\\ \text{By inspection I can see that }f=2\\ f(x)+2=(x+1)^3(dx^2+ex+2)\\ f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\ \text{The x term will be }6x+ex=(6+e)x\\ \text{ but the coefficient of x must be 0 so e=-6}\\~\\ \text{The x squared term will be }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\ \text{ but the coefficient of x squared must be 0 so d=12}\\ so\\ f(x)+2=(x+1)^3(dx^2+ex+f)\\ f(x)+2=(x+1)^3(12x^2-6x+2)\\ \text{expand etc}$$

The logic is good but you need to check for careless errors.

LaTex:

f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But  also}\\~\\
\text{By inspection I can see that   }f=2\\
f(x)+2=(x+1)^3(dx^2+ex+2)\\
f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\
\text{The x term will be   }6x+ex=(6+e)x\\
\text{ but the coefficient of x must be 0 so e=-6}\\~\\
\text{The x squared term will be   }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\
\text{ but the coefficient of x squared must be 0 so d=12}\\
so\\
f(x)+2=(x+1)^3(dx^2+ex+f)\\
f(x)+2=(x+1)^3(12x^2-6x+2)\\
\text{expand etc}

May 7, 2022
edited by Melody  May 7, 2022