Solving simultaneous equations with multiple variables. If \(\displaystyle{\left({x}+{a}\right)}^{{2}}\)

markush35q

markush35q

Answered question

2022-04-03

Solving simultaneous equations with multiple variables.
If (x+a)2 is a factor of x3+6px+k, show that k+2a3=0
I've tried different ways to solve this, but is it even possible?

Answer & Explanation

microsgopx6z7

microsgopx6z7

Beginner2022-04-04Added 14 answers

Let p(x) denote the 3rd degree polynomial.
Your condition means that -a is a double root of p, therefore it is also a root of the derivative p(x)=3x2+6p.
As a consequence, we have:
{p(a)=a36pa+k=0p'(a)=3a2+6p=0
from which you easily derive your constraint.

Kamora Campbell

Kamora Campbell

Beginner2022-04-05Added 13 answers

Since (x+a)2 divides x3+6px+k exactly, the cubic polynomial must be of the form (x+a)2(xb). Since it has no x2 term, the sum of the roots must be zero. So (a)+(a)+b=0, hence b=2a.
Therefore we have x3+6px+k=(x+a)2(x2a). The constant term k must be equal to the product of the constant terms on the right. So k=(a)(a)(2a)=2a3.

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