# problem

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x^3 - 4x^2 - 25 has a factor (x-a). Find the value of a.

I've tried two methods for this. The first one basically went

f(a)= a^3 - 4x^2 - 25 = 0

SQRT(a^3 - 4x^2 - 25) = 0

a^(3/2) - 2x - 5 = 0

dunno where to go from there.

The other method i tried was something like

x^3 - 4x^2 - 25 = (x-a)(bx^2 + cx + d)

= bx^3 + cx^2 + dx - abx^2 -acx -ad

= bx^3 + (c-ab)x^2 + (d-ac)x -ad

Equate coeffecients:

b = 1

c-ab = -4

c-a(1) = -4

c-a = -4

d-ac = 0

I've then tried various combinations to try to get an answer for any of the letters but to no avail.

The answers is apparently a=5 but afaik a^(3/2) - 2x - 5 = 0 (from first attempt) is correct and when i substitute a=5 into that it doesn't =0.

Any help appreciated, evan if it's just to confirm that the text book is wrong.

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Hi,

It is easily verified that x = 5,is a solution to f(x) = 0, right ?

So since it is a polnomial the only thing you will need to proof is that you can write every polynonomial as the product of (x - r), where r is a root of the polynomial and you are finished.

By the way there are explicit formulas for finding the roots of third degree polynomials (The formulaes of Cardano).

So try proving that if p(x) is a polynomial of degree n and x_1,...x_n are its roots (counting multiplicities) that p(x) = prod_i=1^n (x - x_i)

Then show a = 5 is a root of your polynomial and you are done

Mandrake

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I think for a question put so simply, you're probably not going to have to prove that every polynomial can be written as a product of its factors.

Basically I just guessed the answer x=5 by just plugging different values in, which is the standard method at A-level in the UK (as far as I'm aware).

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Yeah that is the most easy way to find roots if you suspect that one of them is "easily" found.

I think i probably misunderstood the nature of the problem.

Mandrake

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