Question

the condition that x³-3px+2q may be divisible by a factor of the form x²+2ax+a² is​

Answer 1

Thus , if p3=q2 then x3−3px+2q may be divisible by the factor of x2+2ax+a2.

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Answer 2

Answer:

$p^{3} =q^{2}$

Step-by-step explanation:

Concept

• Any mathematical statement containing numbers, variables, and an arithmetic operation between them is referred to as an expression or algebraic expression.

Given

$x^{3} -3px+2q$

Find

Condition when given expression divisible by $x^{2} +2ax+a^{2}$

Solution

As $x^{3}-3 p x+2 q$ is divisible by $x^{2}+2 a x+a^{2}$ that is $(x+a)^{2}$ which means the cubic equation must have three roots .

Let the three roots be $\alpha, \beta, \gamma$.

Thus $\alpha=-\mathrm{a}$

$beta=-a$

Thus sum of three roots of cubic equation will be $\alpha+\beta+\gamma=(-a)+(-a)+\gamma= -\frac{\left.x^{2} \text { (coefficient }\right)}{x^{3}(\text { coefficient })}$

$-2 a+\gamma=0$

Thus,$\alpha \beta+\beta \gamma+\gamma \alpha=\frac{x \text { (coefficient) }}{x^{3} \text { (coefficient) }}$

Therefore,

$a^{2}-a \gamma-a \gamma=-3 p$

$a^{2}-2 a \gamma=-3 p$

$a^{2}-2 a \cdot 2 a=-3 p$

$a^{2}-4 a^{2}=-3 p$

$-3 a^{2}=-3 p$

$\mathrm{p}=\mathrm{a}^{2}$

$\mathrm{p}^{3}=\mathrm{a}^{6}$

Product of three roots of cubic equation will be$\alpha \cdot \beta \cdot \gamma=(-a) \cdot(-a) \cdot \gamma=\frac{\text { constant (coefficient) }}{\mathrm{x}^{2} \text { (coefficient) }}$

Thus, $a^{2} \gamma=-2 q$

$a^{2} \cdot 2 a=-2 q$

$2 a^{3}=-2 q$

$q=-a^{3}$

squaring both side,

$q^{2}=a^{6}$

So if $p^{3} =q^{2}$ then $x^{3} -3px+2q$ will be divisible by a factor of the form $x^{2} +2ax+a^{2}$

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