# Polynomials - Remainder thereom + factor thereom

Write $p(x) = x^4 + 4x^3 - 14x^2 - 36x + 45$ as a product of its factors.

My solution so far:

$p(3) = 0$ therefore it's a factor $$(x^4 + 4x^3 - 14x^2 - 36x + 45)\big/(x-3) ~=~ x^3 + 7x^2 + 7x - 15$$ and i am upto $(x-3)( x^3 + 7x^2 + 7x - 15)$... i cant figure out how to finalize the answer.

THE SOLUTION IS : $(x+3)(x-3)(x+5)(x-1)$

ANSWER: Write p(x)=x4+4x3−14x2−36x+45 as a product of its factors.

My solution so far:

p(3)=0 therefore it's a factor (x4+4x3−14x2−36x+45)/(x−3) = x3+7x2+7x−15 and i am upto (x−3)(x3+7x2+7x−15)... i cant figure out how to finalize the answer.

p(-3) is a factor of (x3 + 7x2 + 7x +15)

....

(x-3)(x+3)(x+5)(x-1)

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please someone !!! :( – MATHSUSER Mar 27 '13 at 9:26

This is a polynomial of 4 degrees. It will have factors of the form

$$(x-a)(x-b)(x-c)(x-d)=0$$

The long method states you multiply the above and compare coefficients with your polynomial and calculate $a,b,c,d$. which will give 3 equation in terms of $a,b,c,d$. The solution is long but definitely will reach there.

The shorter method is substitution where, you assume one value of x that satisfy the polynomial.

Now,

$p(x) = x^4 + 4x^3 - 14x^2 - 36x + 45$

Substituting x=1 gives $1+4-14-36+45=0$ So, x-1 is a solution.

Now divide $p(x)$ by$(x-1)$ using polynomial long division and subsequently reduce the polynomial by 1 degree, and repeat till you get your factors..

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You basically want to factor $x^3+7x^2+7x-15$ Let's hope that there are rational roots. If they are rational, they must be integral, looking at the coefficient of $x^3$ Also, they must be factors of $15$, We try 1 as a root. It works. Now we have a quadratic which can be factored by completing the square.

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i sort of understand what you are trying to imply, but not clearly. can you please expand. – MATHSUSER Mar 27 '13 at 9:28
@ Ishan Banerjee can you please expand. i really appreciate that, – MATHSUSER Mar 27 '13 at 9:32
Just the way you have concluded $3$ is a root and hence $x-3$ is a factor, conclude for $1$ also. – Macavity Mar 27 '13 at 9:39
Whenever a polynomial is given, always start with $1$ and $-1$.
If sum of coefficients of the polynomial is zero, then $1$ is a root and $(x-1)$ is a factor.
If sum of even coefficients equals sum of odd coefficients, the $-1$ is a root and $(x+1)$ is a factor.