Answer to Question #104444 in Algebra for maria

Question

Find the value of a for which the polynomial x^5-ax²-ax+1has −1 as a root with multiplicity at least 2.

Answer 1

ANSWER: a=-5

EXPLANATION.

$P_5(x):=x^5-ax^2-ax+1=(x^5+1)-ax(x+1)$ $\quad { { P }_{ 4 }(x)=\frac { { P }_{ 5 }(x) }{ x+1 } ={ x }^{ 4 }{ -x }^{ 3\quad }+{ x }^{ 2 } }-x+1-ax$ . $P_4(-1)=0$ , since $x=-1$ is the root of the second multiplicity of the polynomial $P_5(x)$ . At the same time, $P_4(-1)=a+5$ .Therefore $a+5=0$ or $a=-5$ .

VERIFICATION: If $a=-5$ , then $P_5(x)=x^5+5x^2+5x+1$ , $\quad \frac { { P }_{ 5 }(x) }{ { \left( x+1 \right) }^{ 2 } } ={ x }^{ 3 }-2{ x }^{ 2 }+3x+1$