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Let $f(x)=x^3-ax^2-bx-3a$ and $g(x)=x^3+(a-2)x^2-bx-3b$. If they have a common quadratic factor, then find the value of $a$ and $b$.

My Attempt

Let $h(x)$ be the common quadratic factor. Then $h(x)$ also the factor of $g(x)-f(x)$, that is $$(2a-2)x^2+(3a-3b)$$

Since $h(x)$ a quadratic factor, then $$(2a-2)x^2+(3a-3b)=k \cdot h(x)$$ Where $k$ is a constant.

But, i don't know how to continue, because there are many possible values for $k$.

Any advice?

Agung Izzul Haq
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2 Answers2

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Use the euclidean algorithm. Your idea of $f(x) -g(x)$ is right, just divide by the leading coefficient.

vonbrand
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  • I am sorry, i don't get it. Could you explain it a bit more? – Agung Izzul Haq Mar 15 '20 at 15:04
  • I have tried based on my understanding about your opinion. I set the first linear remainder to zero, since the gcd of $f(x)$ and $g(x)$ must have degree 2 or 3. Does it correct? – Agung Izzul Haq Mar 16 '20 at 01:59
  • @AgungIzzulHaq, your polynomials are monic, so $f(x) - g(x)$ is the first step in the euclidean algorithm. Next step would be to divide e.g. $f(x)$ by $f(x) - g(x)$ and keep the remainder (if not zero). If they really have a quadratic common factor, the remainder of the second step will be zero. – vonbrand Mar 17 '20 at 03:10
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Since $(2a-2)x^2+(3a-3b)$ is a common factor and the common factor is said to be a quadratic, the (monic) common factor must look like $h(x) = x^2+\dfrac 32 \cdot \dfrac{a-b}{a-1}$.

This is ugly. So let's try and avoid the direction that this is taking us. We can assume that $h(x) = x^2 - \alpha$ where $\alpha = -\dfrac 32 \cdot \dfrac{a-b}{a-1}$ and, for some $u$ and for some $v$

\begin{align} f(x) &= (x-u)(x^2-\alpha) \\ x^3-ax^2-bx-3a &= x^3 - ux^2 - \alpha x +\alpha u \\ u &= a \\ \alpha &= b \\ \alpha u &= -3a \end{align}

\begin{align} g(x) &= (x-v)(x^2-\alpha) \\ x^3+(a-2)x^2-bx-3b &= x^3 - vx^2 - \alpha x +\alpha v \\ v &= 2-a \\ \alpha &= b \\ \alpha v &= -3b \end{align}

Since $\alpha = b$, then $\alpha v = -3b \implies b v = -3b \implies b(v+3)=0$.

So, either $b=0$ or $v=-3$

If $b=0$, then we must also have $a=0$.

In which case,

\begin{align} f(x) &=x^3 \\ g(x) &=x^3-2x^2 \\ h(x) &=x^2 \end{align}

If $v=-3$, then $a=5$, $u=5$, $\alpha = -3$, and $b=-3$.

In which case

\begin{align} f(x) &= x^3-5x^2+3x-15 \\ g(x) &= x^3+3x^2+3x+9 \\ h(x) &= x^2+3 \end{align}

Steven Alexis Gregory
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