Polyomials question (1 Viewer)

Aviator_13 · Jan 19, 2019

Need help with this polynomials question. Ive attached a photo to this post (dont really know how to attach it properly but i think this works).

If you could explain your thought process and your steps it would be greatly appreciated.

jathu123 · Jan 19, 2019

$\noindent \text{Begin by writing the long division statement:} \\ P(x)=(x^2-5)Q(x)+x+4 \\ \text{We also require P(-x), and we can find this out from the expression above:} \\P(-x)=((-x)^2-5)Q(-x)-x+4 \\ \\ \text{Adding both these expressions, we get:} \\ P(x)+P(-x)=(x^2-5)Q(x)+x+4+(x^2-5)Q(-x)-x+4 \\ = (x^2-5)\left ( Q(x)+Q(-x) \right )+8 \\ \\ \text{This means that when P(x)+P(-x) is divided by} \ x^2-5, \text{the remainder is 8 (and the quotient is} \ Q(x)+Q(-x)) \\ \text{So our answer is just 8.}$

Moderator note: LaTeX input has been adjusted to make the equation valid and visible.

Aviator_13 · Jan 19, 2019

Thanks so much!!

kkk579 · Jan 7, 2023

can someone pls solve this for me because i don’t have the answer

Average Boreduser · Jan 7, 2023

I mean I'll try i guess

kkk579 · Jan 7, 2023

Pethmin said:
I mean I'll try i guess

thankss

Average Boreduser · Jan 7, 2023

Are these polynomial identities?

kkk579 · Jan 7, 2023

Pethmin said:
This requires polynomial identities which is a 3U topic I have not cover from du tuition
I think main thing we're looking for here is p(x) and everything else is smooth sailing

Ohh okay ic then is anyone else able to solve it?
I think i haven’t learnt it either

Average Boreduser · Jan 7, 2023

kkk579 · Jan 7, 2023

Pethmin said:
$P(x)=(x^2 -5)(something) + (x+4)$

that’s what i did

kkk579 · Jan 7, 2023

can some math genius pls
help me

Average Boreduser · Jan 7, 2023

is q(x)=x+sqrt5?

jimmysmith560 · Jan 7, 2023

@kkk579 I just fixed the LaTeX input in the solution provided by the user in the thread that you were talking about. The solution is as follows:

$\noindent \text{Begin by writing the long division statement:} \\ P(x)=(x^2-5)Q(x)+x+4 \\ \text{We also require P(-x), and we can find this out from the expression above:} \\P(-x)=((-x)^2-5)Q(-x)-x+4 \\ \\ \text{Adding both these expressions, we get:} \\ P(x)+P(-x)=(x^2-5)Q(x)+x+4+(x^2-5)Q(-x)-x+4 \\ = (x^2-5)\left ( Q(x)+Q(-x) \right )+8 \\ \\ \text{This means that when P(x)+P(-x) is divided by} \ x^2-5, \text{the remainder is 8 (and the quotient is} \ Q(x)+Q(-x)) \\ \text{So our answer is just 8.}$

Fun fact: this question appeared in the Cambridge Mathematics Extension 1 year 11 textbook.

kkk579 · Jan 7, 2023

jimmysmith560 said:
@kkk579 I just fixed the LaTeX input in the solution provided by the user in the thread that you were talking about. The solution is as follows:

$\noindent \text{Begin by writing the long division statement:} \\ P(x)=(x^2-5)Q(x)+x+4 \\ \text{We also require P(-x), and we can find this out from the expression above:} \\P(-x)=((-x)^2-5)Q(-x)-x+4 \\ \\ \text{Adding both these expressions, we get:} \\ P(x)+P(-x)=(x^2-5)Q(x)+x+4+(x^2-5)Q(-x)-x+4 \\ = (x^2-5)\left ( Q(x)+Q(-x) \right )+8 \\ \\ \text{This means that when P(x)+P(-x) is divided by} \ x^2-5, \text{the remainder is 8 (and the quotient is} \ Q(x)+Q(-x)) \\ \text{So our answer is just 8.}$

Fun fact: this question appeared in the Cambridge Mathematics Extension 1 year 11 textbook.

Also, please avoid further derailing this thread. If you wish to discuss maths questions, feel free to post in the maths forums

Okay thank you, i also got the same answer of 8

Average Boreduser · Jan 7, 2023

Ohh so they cancel out. thanks jim,

kkk579 · Jan 7, 2023

jimmysmith560 said:
@kkk579 I just fixed the LaTeX input in the solution provided by the user in the thread that you were talking about. The solution is as follows:

$\noindent \text{Begin by writing the long division statement:} \\ P(x)=(x^2-5)Q(x)+x+4 \\ \text{We also require P(-x), and we can find this out from the expression above:} \\P(-x)=((-x)^2-5)Q(-x)-x+4 \\ \\ \text{Adding both these expressions, we get:} \\ P(x)+P(-x)=(x^2-5)Q(x)+x+4+(x^2-5)Q(-x)-x+4 \\ = (x^2-5)\left ( Q(x)+Q(-x) \right )+8 \\ \\ \text{This means that when P(x)+P(-x) is divided by} \ x^2-5, \text{the remainder is 8 (and the quotient is} \ Q(x)+Q(-x)) \\ \text{So our answer is just 8.}$

Fun fact: this question appeared in the Cambridge Mathematics Extension 1 year 11 textbook.

Also, please avoid further derailing this thread. If you guys wish to discuss maths questions, feel free to post in the maths forums.

But for the p(-x) part, isn’t q(x) just left like that, why would you replace x with -x because q(x) is the quotient not the actual equation if u get what i mean

Average Boreduser · Jan 7, 2023

kkk579 said:
But for the p(-x) part, isn’t q(x) just left like that, why would you replace x with -x because q(x) is the quotient not the actual equation if u get what i mean

I think its bc either it would cancel out e.g: if Q(x)=x, Q(-x)=-x----> Q(x)+Q(-x)=x-x=0 (cancels out)

Hence, then it would be (x^2-5)(0)+8=8

kkk579 · Jan 7, 2023

Pethmin said:
I think its bc either it would cancel out e.g: if Q(x)=x, Q(-x)=-x----> Q(x)+Q(-x)=x-x=0 (cancels out)

Hence, then it would be (x^2-5)(0)+8=8

idk im still confused lol maybe im overthinking it

kkk579 · Jan 7, 2023

when it says p(-x)=____Q(-x)_____
i thought that you would only sub -x into ‘x^2-5’and ‘x’

kkk579 · Jan 7, 2023

@jimmysmith560 Do u mind explaining

Polyomials question (1 Viewer)

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