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#### wishmaster

##### Active member

- Oct 11, 2013

- 211

\(\displaystyle x^3-9x^2+27x-27\)

If possible,without Horners algorithm. Thank you!

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- Oct 11, 2013

- 211

\(\displaystyle x^3-9x^2+27x-27\)

If possible,without Horners algorithm. Thank you!

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- Oct 11, 2013

- 211

Yes MArk,but what is the next step?Perhaps if you write it as:

\(\displaystyle x^3+3x^2(-3)+3x(-3)^2+(-3)^3\)

Does this look like a familiar expansion?

What should i do with \(\displaystyle 3^n\) terms?

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Consider that:Yes MArk,but what is the next step?

What should i do with \(\displaystyle 3^n\) terms?

\(\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3\)

What are $a$ and $b$ in the case of the given expression?

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- Oct 11, 2013

- 211

$a$ is $x$ and $b$ is $-3$ ?Consider that:

\(\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3\)

What are $a$ and $b$ in the case of the given expression?

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Yes, that's correct!$a$ is $x$ and $b$ is $-3$ ?

So, what is the factored form?

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- Oct 11, 2013

- 211

\(\displaystyle (x-3)^3\) so the root of the polynomial is $3$.Yes, that's correct!

So, what is the factored form?

I cant switch my brains to mathematical thinking.......i get stucked by easy problems like this,and that is no good......

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It comes with practice...you will find the more practice and experience you have, the more quickly you recognize patterns you have seen before.\(\displaystyle (x-3)^3\) so the root of the polynomial is $3$.

I cant switch my brains to mathematical thinking.......i get stucked by easy problems like this,and that is no good......

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- #9

- Oct 11, 2013

- 211

Yes,i think so...actually,i have no books,or something to learn,only online help. So here on the forum, and especially you MArk,are very helpfull for me......It comes with practice...you will find the more practice and experience you have, the more quickly you recognize patterns you have seen before.

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We are glad to help here at MHB.Yes,i think so...actually,i have no books,or something to learn,only online help. So here on the forum, and especially you MArk,are very helpfull for me......

Also, I forgot to mention that your factored form is correct.

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- #11

- Oct 11, 2013

- 211

Thank you!We are glad to help here at MHB.

Also, I forgot to mention that your factored form is correct.

Yes,its correct,but im not happy beacuse i didnt come alone to the solution.....

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Try another method then. Pretend you don't know the answer, and see if you can instead apply the rational roots theorem.Thank you!

Yes,its correct,but im not happy beacuse i didnt come alone to the solution.....

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- #13

- Oct 11, 2013

- 211

Wish i could know the other method.......Try another method then. Pretend you don't know the answer, and see if you can instead apply the rational roots theorem.

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Here is an article on it:Wish i could know the other method.......

Rational root theorem - Wikipedia, the free encyclopedia

This theorem tells us that if the given polynomial has rational roots, it will come from the list:

\(\displaystyle \pm\left(1,3,9,27 \right)\)

So we let:

\(\displaystyle f(x)=x^3-9x^2+27x-27\)

and when we find a number $k$ from the list such that:

\(\displaystyle f(k)=0\)

then we know $x-k$ is a factor, and we may use polynomial or synthetic division to get:

\(\displaystyle f(x)=(x-k)P(x)\)

And then we see if we can then further factor $P(x)$.