The polynomial \(2x^4+ax^3+4x^2+bx-3\) is denoted by \(p(x)\).
It is given that \((x^2+x+1)\) is a factor of \(p(x)\).
(a) Find the values of \(a\) and \(b\).
(b) Hence, show that \((x+3)\) is a factor of \(p(x)\).
Solution
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(a) Since \((x^2+x+1)\) is a factor of \(p(x)\), divide \(p(x)\) by \(x^2+x+1\).
\[
p(x)=2x^4+ax^3+4x^2+bx-3.
\]
Using polynomial division, the remainder is
\[
(b-2)x+(a-7).
\]
For \(x^2+x+1\) to be a factor, the remainder must be zero.
Therefore
\[
b-2=0
\]
and
\[
a-7=0.
\]
So
\[
a=7,\quad b=2.
\]
Answer: \(a=7\), \(b=2\).
(b) With \(a=7\) and \(b=2\),
\[
p(x)=2x^4+7x^3+4x^2+2x-3.
\]
Substitute \(x=-3\):
\[
p(-3)=2(-3)^4+7(-3)^3+4(-3)^2+2(-3)-3.
\]
\[
=162-189+36-6-3=0.
\]
Since \(p(-3)=0\), by the factor theorem, \((x+3)\) is a factor of \(p(x)\).
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