Submitted by nkoparan on Sun, 08/06/2017 - 12:42
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For the step question where did we get the 175 and 1323? Also why did we use the cases where x=-1 and x=1?
Submitted by nkoparan on Sun, 08/06/2017 - 12:42
For the step question where did we get the 175 and 1323? Also why did we use the cases where x=-1 and x=1?
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175 and 1323 are found...
Hi !
(Apologies if there are mistakes in the mathematical volcabulary, I'm not a British student so I don't quite know the english Maths vocab)
The step question is in two part here : first there is a general case for a polynomial function, for which you need to find three equalities :
k(1) * k(2) * ... * k(n) = ...
( k(1) + 1 ) * ( k(2) + 1 ) * ... ( k(n) + 1 ) = ...
( k(1) - 1 ) *( k(2) - 1 ) * ... ( k(n) - 1 ) = ...
Once you have done this, you are given a polynom of the 4th degree to study - you just have to apply the former results in this particular case to get the three equations.
Equations
You are told that:
\[
f(x) = x^n + a_1x^{n-1} +a_2x^{n-2}+\cdots+a_n = (x+k_1)(x+k_2)\cdots(x+k_n).
\]
Substituting $x=0$ into this gives:
\[
a_n = k_1k_2 \cdots k_n.
\]
Substituting $x=1$ gives:
\[
1 + a_1 +a_2+\cdots+a_n = (1+k_1)(1+k_2)\cdots(1+k_n).
\]
Substituting $x=-1$ gives:
\[
(-1)^n + a_1 \times (-1)^{n-1} +a_2 \times (-1)^{n-2}+\cdots+a_n = (-1+k_1)(-1+k_2)\cdots(-1+k_n).
\]
Now you can consider the special case with the quartic (hence $n=4$) given:
\[
k_1k_2k_3k_4=a_4=576
\]
\[
(1+k_1)(1+k_2)(1+k_3)(1+k_4)=1 + a_1 +a_2+a_3+a_4 \\
\qquad \qquad \qquad \qquad \qquad =1+22+172+552+576=1323
\]
\[
(k_1-1)(k_2-1)(k_3-1)(k_4-1)=1 - a_1 +a_2-a_3+a_4 \\
\qquad \qquad \qquad \qquad \qquad =1-22+172-552+576=175
\]