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Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.

The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).

Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.

The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).

Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.

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Two ways:

First, look at the coefficients: $1, 4, 6, 4, 1$ with alternating signs. If you know the binomial theorem, you can easily associate this to the expansion of $(x-1)^4$. You'd have to be pretty darn clever.

Second, you can just find the roots via synthetic division. You'll see that you'll have $x=1$ as the only root with multiplicity $4$. This corresponds to the factors $(x-1)(x-1)(x-1)(x-1)$.

You can even check that $(x-1)^4$ yields the above expansion by multiplying.

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  • $\begingroup$ Why does the whole problem not factor? $\endgroup$ – William Zlacki Jul 28 '16 at 20:14
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    $\begingroup$ It does factor, but you would have to be very, very slick to see it immediately. I've given the factorization in two ways here. $\endgroup$ – Sean Roberson Jul 28 '16 at 20:16
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You can factor it as

$$x^{ 4 }-4x^{ 3 }+6x^{ 2 }-4x+1={ x }^{ 4 }-2{ x }^{ 3 }+{ x }^{ 2 }-2{ x }^{ 3 }+4{ x }^{ 2 }-2x+{ x }^{ 2 }-2x+1={ x }^{ 2 }\left( { x }^{ 2 }-2x+1 \right) -2x\left( { x }^{ 2 }-2x+1 \right) +{ x }^{ 2 }-2x+1=\\ =\left( { x }^{ 2 }-2x+1 \right) \left( { x }^{ 2 }-2x+1 \right) ={ \left( x-1 \right) }^{ 2 }{ \left( x-1 \right) }^{ 2 }={ \left( x-1 \right) }^{ 4 }\\ $$

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The rational root theorem tells you that any rational root $\frac{p}{q}$ will have $p\mid 1$ and $q \mid 1$ so $\frac{p}{q} = \pm 1$, you can then verify that only $x=1$ is a root and be done.

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    $\begingroup$ Doesn't tell you that x=1 is a multiple root. There could be three irrational and/or complex roots. $\endgroup$ – fleablood Jul 28 '16 at 21:12
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Knowing that x=1 is a solution we know that $(x-1) $ is a factor so we force it to factor out.

$x^4-4x^3+6x^2-4x+1=$

$(x-1)x^3-3x^3 +6x^2-4x+1=$

$(x-1)x^3-(x-1)3x^2+3x^2-4x+1=$

$(x-1)x^3-(x-1)3x^2 +(x-1)3x -x +1=$

$(x-1)x^3-(x-1)3x^2+(x-1)3x-(x-1)=$

$(x-1)[x^3-3x^2+3x-1]=$

Now, we may not know that 1 is a double root yet. But if $x=1$ then $x^3-3x^2+3x-1=0$ so we do know that $x-1$ will factor out again. So we force it.

$(x-1)[(x-1)x^2-2x^2+3x-1]=$

$(x-1)[(x-1)x^2-(x-1)2x+x-1]=$

$(x-1)[(x-1)x^2-(x-1)2x+(x-1)]=$

$(x-1)(x-1)[x^2-2x+1]=$

I'll pretend it's not obvious at this point.

We test $x =1$ so $x^2-2x+1=0$ so we know $(x-1) $ must factor a third time. We force it out but by this point it should look like something very clear.

$(x-1)^2 [(x-1)x-x+1]=$

$(x-1)^2 [(x-1)x-(x-1)]=$

$(x-1)^2 (x-1)[x-1]=$

$(x-1)^4$

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$$ \begin{align} n^4 - 4n^3 + 6n^2 - 4n + 1 = 0\\ n^4 - n^3 - 3n^3 + 3n^2 + 3n^2 - 3n - n + 1 = 0\\ n^3(n - 1) - 3n^2(n - 1) + 3n(n - 1) - 1(n-1) = 0\\ (n - 1)(n^3 - 3n^2 + 3n - 1) = 0\\ (n - 1)(n^3 - n^2 -2n^2 + 2n + n - 1) = 0\\ (n - 1)[n^2(n - 1) - 2n(n - 1) + 1(n - 1)] = 0\\ (n - 1)(n - 1)(n^2 - 2n + 1) = 0\\ (n - 1)(n - 1)(n - 1)^2 = 0\\ (n - 1)^4 = 0\\ n - 1 = 0\\ \therefore n = 1 \end{align} $$

enter image description here

Hope you get it.....

.....--))

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Other way $$x^2\left(x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}\right)=0$$ $x=0$ is not a solution of this equation , therefore $$x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}=0$$ we have $$\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0$$ set $t=x+\frac{1}{x}$, thus $$t^2-4t+4=(t-2)^2=0$$ Hence $$t=x+\frac{1}{x}=2$$ i.e. $x=1$

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    $\begingroup$ Your first equation, why isn't $x=0$ a solution, due to $x^2=0$? $\endgroup$ – JDoeDoe Jul 28 '16 at 21:14
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    $\begingroup$ If x = 0 then 4/x and 1/x^2 are undefined. In fact we can't do this factoring unless we state x not equal 0. But as x=0 yields 6 not equal to 0 we know 0 is not a solution and we can claim x isn't 0. $\endgroup$ – fleablood Jul 28 '16 at 21:19

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