Polynomial

Uchiha Sasuke

28/03/2019 at 10:52

Given a polynomial f(x) such that $x^3+2x^2\left(4y-1\right)-4xy^2-f\left(x\right)=-5^3+4xy\left(2x-y\right)$. The value of $f\left(\dfrac{f\left(-5\right)}{5^2}\right)$=...

Polynomial

LOL

24/07/2018 at 13:54

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When the polynomial f (x) is divisible by x + 2 and x² + 1, it leaves the remainders 7 and x + 4 respectively. What is the remainder when f (x) is divided by (x + 2)(x² + 4)

Polynomial

Huỳnh Anh Phương 28/07/2018 at 03:10

you have a wrong question

it's divisible when it's remander 0.

LOL

24/07/2018 at 13:52

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When the polynomial f (x) is divisible by x + 2 and x² + 1, it leaves the reminders 7 and x + 4 respectively. What is the remainder when f (x) is divided by (x + 2)(x² + 1)

Polynomial

Lê Anh Duy 24/07/2018 at 17:08

I think you should say: f(x) is divided by ..., because f(x) is divisible when it doesn't have any remainder.

Suppose that the polynomial $f\left(x\right)=2x^5-9x^3+2x^2+9x-3$ has 5 solutions x₁ ; x₂; x_3;x₄; x₅. The other polynomial $k\left(x\right)=x^2-4$_.Find the value of $P=k\left(x_1\right)\times k\left(x_2\right)\times k\left(x_3\right)\times k\left(x_4\right)\times k\left(x_5\right)$.

Polynomial

Uchiha Sasuke

25/05/2018 at 08:52

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Suppose that the polynomial $f\left(x\right)=x^5-x^4-4x^3+2x^2+4x+1$ has 5 solutions $x_1;x_2;x_3;x_4;x_5$.The other polynomial $K (x) = x^{2} - 4$ . Find the value of $P = K (X_{1}) \times K (X_{2}) \times K (X_{3}) \times K (X_{4}) \times K (X_{5})$

Polynomial

Uchiha Sasuke

25/05/2018 at 08:46

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Suppose that the polynomial $f\left(x\right)=2x^5-9x^3+2x^2+9x-3$ has 5 solutions $x_1;x_2;x_3;x_4;x_5$. The other polynomial $K\left(x\right)=x^2-4$. Find the value of $P=K\left(X_1\right)\times K\left(X_2\right)\times K\left(X_3\right)\times K\left(X_4\right)\times K\left(X_5\right)$

Polynomial

Uchiha Sasuke

25/05/2018 at 08:39

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Let P(x) be the polynomial given by $P\left(x\right)=\left(2+x+2x^3\right)^{15}$. Suppose that $P\left(x\right)=a_0+a_1x+a_2x^2+...+a_{45}x^{45}$.The value of $S=a_1-a_2+a_3-a_4+...-a_{44}+a_{45}$.

Polynomial

Uchiha Sasuke 26/05/2018 at 01:19

Thank you very much
Dao Trong Luan Coodinator 25/05/2018 at 12:30

P(x) = a₀ + a₁x + a₂x² + ... + a₄₅x⁴⁵

=> P(1) = a₀ + a₁ + a₂ + ... + a₄₅ = (2 + 1 + 2)¹⁵ = 5¹⁵

=> P(-1) = a₀ - a₁ + a₂ - ... - a₄₅ = (2 - 1 - 2)¹⁵ = (-1)¹⁵

==> P(1) - P(-1) = 2a₁ + 2a₃ + ... + 2a₄₅ = 5¹⁵ + 1

P(1) + P(-1) = 2a₀ + 2a₂ + ... + 2a₄₄ = 5¹⁵ - 1

=> (a₁ + a₃ + ... + a₄₅) - (a₀ + a₂ + ... + a₄₄) = $\dfrac{5^{15}+1-5^{15}+1}{2}=1$

=> S = -a₀ + a₁ - a₂ + .... - a₄₄ + a₄₅ = 1

P/s: Not S in the topic

Uchiha Sasuke

26/01/2018 at 07:48

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Given a third degree polynomial P (x). Find the coefficient of x³ of P (x) such that P(0) = 10; P(1) = 12; P(2) = 4; P(3) = 1

Polynomial

tranthuydung

09/03/2017 at 18:52

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Let P(x) be a polynomial of degree 2015.

Suppose P(n)=$\dfrac{n}{n+1}$ for all n = 0, 1 , 2 ,..., 2015.

The value of P(2016) is ...

Polynomial

John 11/03/2017 at 09:07

Let $Q\left(x\right)=\left(x+1\right)P\left(x\right)-x$, (*)

since $P\left(n\right)=\dfrac{n}{n+1}$ we infer:

$Q\left(n\right)=\left(n+1\right)P\left(n\right)-n=0$ for $n=0,1,..,2015$.

Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

  $Q\left(x\right)=a\left(x-0\right)\left(x-1\right)...\left(x-2015\right)$ (**)

In other hand, in (*) we set x = -1 then  $Q\left(-1\right)=1$. And replace to (**) we have:

  $1=Q\left(-1\right)=a\left(-1\right)\left(-2\right)...\left(-2016\right)$

$\Rightarrow a=\dfrac{1}{2016!}$

Finally $Q\left(x\right)=\dfrac{1}{2016!}\left(x-0\right)\left(x-1\right)...\left(x-2015\right)$

$\Rightarrow Q\left(2016\right)=\dfrac{1}{2016!}2016.2015...1=1$

from (*) $Q\left(2016\right)=\left(2016+1\right)P\left(2016\right)-2016$

$\Rightarrow\left(2016+1\right)P\left(2016\right)-2016=1$

$\Rightarrow P\left(2016\right)=\dfrac{2017}{2017}=1$
tranthuydung selected this answer.
FA KAKALOTS 28/01/2018 at 22:05

Let Q(x)=(x+1)P(x)−x

, (*)

since P(n)=nn+1

we infer:

Q(n)=(n+1)P(n)−n=0

for n=0,1,..,2015

.

Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

  Q(x)=a(x−0)(x−1)...(x−2015)

(**)

In other hand, in (*) we set x = -1 then  Q(−1)=1

. And replace to (**) we have:

  1=Q(−1)=a(−1)(−2)...(−2016)

⇒a=12016!

Finally Q(x)=12016!(x−0)(x−1)...(x−2015)

⇒Q(2016)=12016!2016.2015...1=1

from (*) Q(2016)=(2016+1)P(2016)−2016

⇒(2016+1)P(2016)−2016=1

⇒P(2016)=20172017=1

Polynomial

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