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

please answer this for me ?

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)


eventually the answer comes to


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please someone !!! :( –  MATHSUSER Mar 27 '13 at 9:26
why not start with p(1)=0 –  Tanuj Wadhwa Mar 27 '13 at 10:54

3 Answers 3

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


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.


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

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
please? please? please? –  MATHSUSER Mar 27 '13 at 9:35
actually never mind i have solved it. –  MATHSUSER Mar 27 '13 at 9:38
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.

Most of the times you come out successful. If not you can always go for synthetic division and find the integer roots. Or in some cases you could also come to a conclusion using the product of roots and sum of roots terms.

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