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A polynomial $p(x)$ with real coefficients is of degree five. The equation $p(x)=0$ has a complex root $2+i$. The graph $y=p(x)$ has the x-axis as a tangent at $(2,0)$ and intersects the cordinate axes at $(-1,0)$ and $(0,4)$. Find $p(x)$ in factorised form with real coefficients.

Firts I found the roots:

$r_1=2+i$, $r_2=2-i$, $r_3=-1$, $r_4=2$ and $r_5=a$, I was not able to find it ( for some reason the anwser considered $r_5=2$, but I was not able to prove it)

$p(x)=B(x+1)(x-2)(x-2+i)(x-2-i)(x-a)$

$p(x)=B(x+1)(x-2)(x^2-4x+5)(x-a)$

$p(0)=4$

then:

$4=B(1)(-2)(+5)(-a)$

$B=\frac{2}{5a}$

Observation: if we consider the value of $a=2$ then the answer is right. However I cannot show that $a=2$

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    $\begingroup$ The graph of $p(x)$ has the $x$-axis as a tangent at $(0,2)$. Tangent means multiple root! $\endgroup$ – Qurultay Apr 2 '18 at 13:12
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    $\begingroup$ You must use that $$p'(0)=2$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 2 '18 at 13:14
  • $\begingroup$ @Dr.SonnhardGraubner You must have meant $p'(2) = 0$. $\endgroup$ – Deepak Apr 2 '18 at 14:08
  • $\begingroup$ Sorry, yes that is what i meant! $\endgroup$ – Dr. Sonnhard Graubner Apr 2 '18 at 14:09
  • $\begingroup$ @Qurultay .& that's all there is to it, as we also have p(0)=4. $\endgroup$ – DanielWainfleet Apr 2 '18 at 14:38
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Let $f(x)=g(x)(x-2)(x-a)$, where $g(x)=B(x+1)(x-2+i)(x-2-i)$. The $x$ axis has slope $0$. Since $f$ is tangent to the $x$ axis at $x=2$, we have

$$f'(2)=0$$ But $$f'(x)=g'(x)(x-2)(x-a)+g(x)(x-a)+g(x)(x-2)$$ so that $$f'(2)=g(2)(2-a)=0$$ Since $g(2)\ne 0$, we must have $a=2$

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As the excellent hint in the comment stated, an $x$-axis tangent means a repeated root. That means that $(x-2)$ is a repeated factor of $p(x)$, i.e that $(x-1)^m, m \geq 2$ is a factor of $p(x)$.

The other obvious factors are $(x+1)$ and $(x-(2+i))(x-(2-i)) = (x^2 - 4x + 5)$, which, when multiplied together, gives you a cubic.

That means that the only remaining factor to achieve a fifth degree polynomial is of degree two, and therefore it must be $(x-2)^2$.

So $p(x) = k(x-2)^2(x+1)(x^2-4x+5)$

Apply the last piece of information, that $p(0) = 4$, which gives $k(4)(1)(5) = 4$ or $k = \frac 15$, so $p(x) = \frac 15(x-2)^2(x+1)(x^2-4x+5)$

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