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

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Solving simultaneous equations with multiple variables.If (x+a)2 is a factor of x3+6px+k, show that k+2a3=0I&#039;ve tried different ways to solve this, but is it even possible?

microsgopx6z7 · Accepted Answer

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&amp;nbsp;p(x)=3x2+6p.As a consequence, we have:{p(a)=−a3−6pa+k=0p&#039;(a)=3a2+6p=0from which you easily derive your constraint.

Kamora Campbell · Accepted Answer

Since (x+a)2 divides x3+6px+k exactly, the cubic polynomial must be of the form (x+a)2(x−b). 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(x−2a). The constant term k must be equal to the product of the constant terms on the right. So k=(a)(a)(−2a)=−2a3.

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

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