Let $a,b,c,d$ be the roots of $x^4-x^3-x^2-1=0$ . Also consider $g(x)=x^6–x^5–x^3–x^2–x$.Find $g(a)+g(b)+g(c)+g(d)$. [closed]

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Let $a,b,c,d$ be the roots of $x^4-x^3-x^2-1=0$.
Also consider
$$
g(x)=x^6–x^5–x^3–x^2–x.
$$
Find $g(a)+g(b)+g(c)+g(d)$.




Please use only algebra.
I have started it using Newton's identity but I got my answer as $5$ but actual answer is 6.



Please help.




algebra-precalculus polynomials




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closed as off-topic by Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist Aug 25 at 19:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
    – Lord Shark the Unknown
    Aug 25 at 3:33






  • 1




    including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
    – Siong Thye Goh
    Aug 25 at 3:37






  • 2




    using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
    – dxiv
    Aug 25 at 3:54











  • Maybe your mistake is around $a+b+c+d=1$.
    – Takahiro Waki
    Aug 25 at 4:25










  • Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
    – dan_fulea
    Aug 25 at 5:24














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Let $a,b,c,d$ be the roots of $x^4-x^3-x^2-1=0$.
Also consider
$$
g(x)=x^6–x^5–x^3–x^2–x.
$$
Find $g(a)+g(b)+g(c)+g(d)$.




Please use only algebra.
I have started it using Newton's identity but I got my answer as $5$ but actual answer is 6.



Please help.




algebra-precalculus polynomials




share|cite|improve this question














closed as off-topic by Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist Aug 25 at 19:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
    – Lord Shark the Unknown
    Aug 25 at 3:33






  • 1




    including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
    – Siong Thye Goh
    Aug 25 at 3:37






  • 2




    using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
    – dxiv
    Aug 25 at 3:54











  • Maybe your mistake is around $a+b+c+d=1$.
    – Takahiro Waki
    Aug 25 at 4:25










  • Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
    – dan_fulea
    Aug 25 at 5:24












up vote
0
down vote

favorite
1









up vote
0
down vote

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1






1






Let $a,b,c,d$ be the roots of $x^4-x^3-x^2-1=0$.
Also consider
$$
g(x)=x^6–x^5–x^3–x^2–x.
$$
Find $g(a)+g(b)+g(c)+g(d)$.




Please use only algebra.
I have started it using Newton's identity but I got my answer as $5$ but actual answer is 6.



Please help.




algebra-precalculus polynomials




share|cite|improve this question















Let $a,b,c,d$ be the roots of $x^4-x^3-x^2-1=0$.
Also consider
$$
g(x)=x^6–x^5–x^3–x^2–x.
$$
Find $g(a)+g(b)+g(c)+g(d)$.




Please use only algebra.
I have started it using Newton's identity but I got my answer as $5$ but actual answer is 6.



Please help.





algebra-precalculus polynomials





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 8:04









Jendrik Stelzner

7,57221037




7,57221037










asked Aug 25 at 3:30









ankit sharma

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42




closed as off-topic by Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist Aug 25 at 19:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist Aug 25 at 19:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Jendrik Stelzner, Brahadeesh, amWhy, Theoretical Economist
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
    – Lord Shark the Unknown
    Aug 25 at 3:33






  • 1




    including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
    – Siong Thye Goh
    Aug 25 at 3:37






  • 2




    using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
    – dxiv
    Aug 25 at 3:54











  • Maybe your mistake is around $a+b+c+d=1$.
    – Takahiro Waki
    Aug 25 at 4:25










  • Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
    – dan_fulea
    Aug 25 at 5:24












  • 1




    Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
    – Lord Shark the Unknown
    Aug 25 at 3:33






  • 1




    including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
    – Siong Thye Goh
    Aug 25 at 3:37






  • 2




    using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
    – dxiv
    Aug 25 at 3:54











  • Maybe your mistake is around $a+b+c+d=1$.
    – Takahiro Waki
    Aug 25 at 4:25










  • Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
    – dan_fulea
    Aug 25 at 5:24







1




1




Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
– Lord Shark the Unknown
Aug 25 at 3:33




Please see math.meta.stackexchange.com/questions/5020 . Also it's good to have your question in the question, not just in the title.
– Lord Shark the Unknown
Aug 25 at 3:33




1




1




including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
– Siong Thye Goh
Aug 25 at 3:37




including your working in the post is better than just telling us what you obtain. We can't tell your mistake at this point of time.
– Siong Thye Goh
Aug 25 at 3:37




2




2




using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
– dxiv
Aug 25 at 3:54





using Newton's identity Before you do that, simplify first. Hint: $,g(a)=a^2-a+1,$.
– dxiv
Aug 25 at 3:54













Maybe your mistake is around $a+b+c+d=1$.
– Takahiro Waki
Aug 25 at 4:25




Maybe your mistake is around $a+b+c+d=1$.
– Takahiro Waki
Aug 25 at 4:25












Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
– dan_fulea
Aug 25 at 5:24




Please show us some steps in the computation, this is the best way to get errors corrected in the own thinking process. For instance, which are the values for $a+b+c+d$, for $ab+ac+ad+bc+bd+cd$, and for $a^2+b^2+c^2+d^2$?
– dan_fulea
Aug 25 at 5:24










3 Answers
3






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Performing long division of polynomials you have
$$x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1)$$
So for $x = a,b,c,d$ you have
$$x^6 - x^5 - x^3 - x^2 - x = x^2 - x + 1$$
Hence your problem reduces to finding $f(a) + f(b) + f(c) + f(d)$ for $f(x) = x^2 - x + 1$. See if you can do that.






share|cite|improve this answer



























    up vote
    1
    down vote













    As @Zarrax has explained,it can be noted that if $x^4-x^3-x^2-1=0$ and $a,b,c,d$ are roots then for the equation $x^6−x^5−x^3−x^2−x$ at $x=a,b,c,d $ the value will be $x^2-x+1$.
    Now $g (a) + g (b)+g (c)+g (d)=a^2+b^2+c^2+d^2-(a+b+c+d)+4$.NOW, $a^2+b^2+c^2+d^2=(a+b+c+d)^2-2 (ab+ac+ad+bc+bd+cd) $. So finally answer is $-1+1-2 (-1)+4=6$






    share|cite|improve this answer



























      up vote
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      Notice that $$g(x)=x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1),$$and $$a^4-a^3-a^2-a=0$$and so on. Then beginalign*g(a)+g(b)+g(c)+g(d)&=a^2+b^2+c^2+d^2-(a+b+c+d)+4\&=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)-(a+b+c+d)+4.endalign*



      By Vieta's theorem, we have $$a+b+c+d=1,~~~ab+ac+ad+bc+bd+cd=-1.$$
      As a result $$g(a)+g(b)+g(c)+g(d)=1^2-2cdot(-1)-1+4=6.$$






      share|cite|improve this answer



























        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

        oldest

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        up vote
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        down vote













        Performing long division of polynomials you have
        $$x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1)$$
        So for $x = a,b,c,d$ you have
        $$x^6 - x^5 - x^3 - x^2 - x = x^2 - x + 1$$
        Hence your problem reduces to finding $f(a) + f(b) + f(c) + f(d)$ for $f(x) = x^2 - x + 1$. See if you can do that.






        share|cite|improve this answer
























          up vote
          1
          down vote













          Performing long division of polynomials you have
          $$x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1)$$
          So for $x = a,b,c,d$ you have
          $$x^6 - x^5 - x^3 - x^2 - x = x^2 - x + 1$$
          Hence your problem reduces to finding $f(a) + f(b) + f(c) + f(d)$ for $f(x) = x^2 - x + 1$. See if you can do that.






          share|cite|improve this answer






















            up vote
            1
            down vote










            up vote
            1
            down vote









            Performing long division of polynomials you have
            $$x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1)$$
            So for $x = a,b,c,d$ you have
            $$x^6 - x^5 - x^3 - x^2 - x = x^2 - x + 1$$
            Hence your problem reduces to finding $f(a) + f(b) + f(c) + f(d)$ for $f(x) = x^2 - x + 1$. See if you can do that.






            share|cite|improve this answer












            Performing long division of polynomials you have
            $$x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1)$$
            So for $x = a,b,c,d$ you have
            $$x^6 - x^5 - x^3 - x^2 - x = x^2 - x + 1$$
            Hence your problem reduces to finding $f(a) + f(b) + f(c) + f(d)$ for $f(x) = x^2 - x + 1$. See if you can do that.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 25 at 4:40









            Zarrax

            34.8k248102




            34.8k248102




















                up vote
                1
                down vote













                As @Zarrax has explained,it can be noted that if $x^4-x^3-x^2-1=0$ and $a,b,c,d$ are roots then for the equation $x^6−x^5−x^3−x^2−x$ at $x=a,b,c,d $ the value will be $x^2-x+1$.
                Now $g (a) + g (b)+g (c)+g (d)=a^2+b^2+c^2+d^2-(a+b+c+d)+4$.NOW, $a^2+b^2+c^2+d^2=(a+b+c+d)^2-2 (ab+ac+ad+bc+bd+cd) $. So finally answer is $-1+1-2 (-1)+4=6$






                share|cite|improve this answer
























                  up vote
                  1
                  down vote













                  As @Zarrax has explained,it can be noted that if $x^4-x^3-x^2-1=0$ and $a,b,c,d$ are roots then for the equation $x^6−x^5−x^3−x^2−x$ at $x=a,b,c,d $ the value will be $x^2-x+1$.
                  Now $g (a) + g (b)+g (c)+g (d)=a^2+b^2+c^2+d^2-(a+b+c+d)+4$.NOW, $a^2+b^2+c^2+d^2=(a+b+c+d)^2-2 (ab+ac+ad+bc+bd+cd) $. So finally answer is $-1+1-2 (-1)+4=6$






                  share|cite|improve this answer






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    As @Zarrax has explained,it can be noted that if $x^4-x^3-x^2-1=0$ and $a,b,c,d$ are roots then for the equation $x^6−x^5−x^3−x^2−x$ at $x=a,b,c,d $ the value will be $x^2-x+1$.
                    Now $g (a) + g (b)+g (c)+g (d)=a^2+b^2+c^2+d^2-(a+b+c+d)+4$.NOW, $a^2+b^2+c^2+d^2=(a+b+c+d)^2-2 (ab+ac+ad+bc+bd+cd) $. So finally answer is $-1+1-2 (-1)+4=6$






                    share|cite|improve this answer












                    As @Zarrax has explained,it can be noted that if $x^4-x^3-x^2-1=0$ and $a,b,c,d$ are roots then for the equation $x^6−x^5−x^3−x^2−x$ at $x=a,b,c,d $ the value will be $x^2-x+1$.
                    Now $g (a) + g (b)+g (c)+g (d)=a^2+b^2+c^2+d^2-(a+b+c+d)+4$.NOW, $a^2+b^2+c^2+d^2=(a+b+c+d)^2-2 (ab+ac+ad+bc+bd+cd) $. So finally answer is $-1+1-2 (-1)+4=6$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 25 at 7:01









                    Jasmine

                    281111




                    281111




















                        up vote
                        1
                        down vote













                        Notice that $$g(x)=x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1),$$and $$a^4-a^3-a^2-a=0$$and so on. Then beginalign*g(a)+g(b)+g(c)+g(d)&=a^2+b^2+c^2+d^2-(a+b+c+d)+4\&=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)-(a+b+c+d)+4.endalign*



                        By Vieta's theorem, we have $$a+b+c+d=1,~~~ab+ac+ad+bc+bd+cd=-1.$$
                        As a result $$g(a)+g(b)+g(c)+g(d)=1^2-2cdot(-1)-1+4=6.$$






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote













                          Notice that $$g(x)=x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1),$$and $$a^4-a^3-a^2-a=0$$and so on. Then beginalign*g(a)+g(b)+g(c)+g(d)&=a^2+b^2+c^2+d^2-(a+b+c+d)+4\&=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)-(a+b+c+d)+4.endalign*



                          By Vieta's theorem, we have $$a+b+c+d=1,~~~ab+ac+ad+bc+bd+cd=-1.$$
                          As a result $$g(a)+g(b)+g(c)+g(d)=1^2-2cdot(-1)-1+4=6.$$






                          share|cite|improve this answer






















                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Notice that $$g(x)=x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1),$$and $$a^4-a^3-a^2-a=0$$and so on. Then beginalign*g(a)+g(b)+g(c)+g(d)&=a^2+b^2+c^2+d^2-(a+b+c+d)+4\&=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)-(a+b+c+d)+4.endalign*



                            By Vieta's theorem, we have $$a+b+c+d=1,~~~ab+ac+ad+bc+bd+cd=-1.$$
                            As a result $$g(a)+g(b)+g(c)+g(d)=1^2-2cdot(-1)-1+4=6.$$






                            share|cite|improve this answer












                            Notice that $$g(x)=x^6 - x^5 - x^3 - x^2 - x = (x^2 + 1)(x^4 - x^3 - x^2 - 1) + (x^2 - x + 1),$$and $$a^4-a^3-a^2-a=0$$and so on. Then beginalign*g(a)+g(b)+g(c)+g(d)&=a^2+b^2+c^2+d^2-(a+b+c+d)+4\&=(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)-(a+b+c+d)+4.endalign*



                            By Vieta's theorem, we have $$a+b+c+d=1,~~~ab+ac+ad+bc+bd+cd=-1.$$
                            As a result $$g(a)+g(b)+g(c)+g(d)=1^2-2cdot(-1)-1+4=6.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 25 at 8:05









                            mengdie1982

                            3,628216




                            3,628216












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