$\large 54x^4-27x^3-90x^2+82x-32$

Find the value of the expression above if $x=\dfrac{2+\sqrt[3]{5}}{3}$ .

The answer is 7.

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I think you've made a mistake while you were dividing the given polynomial by the one that you created from expression
$x=\frac{2+\sqrt[3]{5}}{3}$
.

I'm saying this because I solved the whole problem and was left with a remainder equal to
$-60x+7$
.

I figured out that mistake was on your end because when I simplified the expression that you finally got in your solution
$2x(27x^3-54x^2+36x-13)+3(27x^3-54x^2+36x-13)+7$
,

it turned out to be
$54x^4-23x^3-84x^2+82x-32$
, while the polynomial in the given problem was
$54x^4-27x^3-90x^2+22x-32$
. So, it means you've solved a polynomial different than the one given in the problem. That means this problem has no solution unless this website allows decimal values for answers to the problems.

Here's the simplification of your expression:

$2x(27x^3-54x^2+36x-13)+3(27x^3-54x^2+36x-13)+7$
$\Rightarrow(54x^4-108x^3+72x^2-26x)+(81x^3-162x^2+108x-39)+7$

$\Rightarrow54x^4-108x^3+81x^3+72x^2-162x^2-26x+108x-39+7$
$\Rightarrow54x^4-27x^3-90x^2+82x-32$

You should be more careful while you're designing questions.

Rakshit Pandey
- 6 years, 11 months ago

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actually i made a mistake in typing the question in place of $22x$ there should be $82x$ . and in your simplification of my last expression you used $-52x^2$ in place of $-54x^2$ . i have editted the question. and it is my first problem that i posted on brilliant .

Shriram Lokhande
- 6 years, 11 months ago

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Thank you for correcting the problem.

Those who previously answered -67 or -68 were marked correct.

@Shriram Lokhande Thanks for pointing out my mistake. I don't like making mistakes while I'm pointing out someone else's. :)

Rakshit Pandey
- 6 years, 11 months ago

There is some confusion you say find the value of the polynomial whose value you already mention as zero .I think you need to make a change there.

Vaibhav Chaturvedi
- 6 years, 11 months ago

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Thanks for notifing didn't spot the mistake , fixed..

Shriram Lokhande
- 6 years, 11 months ago

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@Shriram Lokhande Nice problem... But you should make sure that the title of the problem doesn't give away the answer. I knew what the answer was going to be even before solving it!

Satvik Golechha
- 6 years, 10 months ago

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first we find a polynomial which has given value of $x$ as a zero $x=\dfrac{2+\sqrt[3]{5}}{3} \Rightarrow (3x-2)^3=5$ on expanding $\Rightarrow 27x^3-54x^2+36x-13$ the polynomial

$54x^4-27x^3-90x^2+82x-32$ $=2x(27x^3-54x^2+36x-13)+3(27x^3-54x^2+36x-13)+7$ hence , we get value as $\boxed{7}$