# how to factorize??

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• Aug 16th 2009, 10:42 PM
louis1234
how to factorize??
hi,
how to factorize this expression?

x^4 + x^2 + 1

actually i know how to factorize, but would like to know there is only one answer to that. I first take out x^2 as common factor to factorize the three terms, then use the identities (a+b)^2 = a^2 + 2ab + b^2 and
a^2 - b^2 = (a - b)(a + b) to factorize

is there any other answer ? will the factorized expression be unique?

2. Another question

I really don't know how to factorize the following expression, after a long time of try

I hope to understand the "techniques" of factorization, instead of just knowing how to do. Thanks all!!!!

acx^2 + (ad + bc)xy - bdy^2

the question is correctly typed, only x and y has power of 2

• Aug 16th 2009, 11:20 PM
red_dog
Isn't it $\displaystyle acx^2+(ad\textcolor{red}{-}bc)xy-bdy^2$

or $\displaystyle acx^2+(ad+bc)xy\textcolor{red}{+}bdy^2$ ?
• Aug 16th 2009, 11:30 PM
ynj
Quote:

Originally Posted by louis1234
hi,
how to factorize this expression?

x^4 + x^2 + 1

actually i know how to factorize, but would like to know there is only one answer to that. I first take out x^2 as common factor to factorize the three terms, then use the identities (a+b)^2 = a^2 + 2ab + b^2 and
a^2 - b^2 = (a - b)(a + b) to factorize

is there any other answer ? will the factorized expression be unique?

2. Another question

I really don't know how to factorize the following expression, after a long time of try

I hope to understand the "techniques" of factorization, instead of just knowing how to do. Thanks all!!!!

acx^2 + (ad + bc)xy - bdy^2

the question is correctly typed, only x and y has power of 2

1.Actually, the "uniqueness"will always hold but the answer will be different. if you factorize it on rational number(parameter is rational), then the answer is yours. But in the real field, $\displaystyle x^2+x+1=(x-\frac{-1-\sqrt{3}i}{2})(x-\frac{-1+\sqrt{3}i}{2})$.
P.S. In the factoralization of polynomial, two factors says to be "equivalent" if $\displaystyle f(x)=dg(x)$for some nonzero constant $\displaystyle d$,not exactly $\displaystyle f(x)=g(x)$!
2.Same question as red_dog