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I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.

Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?

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closed as off-topic by Ali Caglayan, colormegone, Mike Pierce, user223391, Claude Leibovici Jun 17 '15 at 8:15

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Since $x^4 + a x^2 + b$ is even (i.e. invariant under $x \to -x$), so is its factorization. If $x^2 + 4 x + 6$ is one factor, the other must be ....

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    $\begingroup$ Top-shelf answer. $\endgroup$ – Orest Bucicovschi Jun 17 '15 at 7:53
  • $\begingroup$ Let us add an ingredient since more complicated answers might mask it: $x^4+ax^2+b=(x^2+4x+6)(x^2-4x+6)$, plugging $x=1$ yields that $1+a+b$ is $(1+4+6)(1-4+6)=(1+6)^2-4^2$ and $a+b=49-16-1=$ $___$. $\endgroup$ – Did Jun 20 '15 at 18:05
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There is some quadratic $x^2 + cx + d$ such that: $$(x^2 + 4x + 6)(x^2 + cx + d) = x^4 + ax^2 + b$$ Multiply it out: $$x^4 + (c + 4)x^3 + (d + 4c + 6)x^2 + (6c + 4d)x + 6d = x^4 + ax^2 + b$$ Equate coefficients:

$$\begin{cases} c + 4 = 0 \\ d + 4c + 6 = a \\ 4d + 6c = 0 \\ 6d = b \end{cases}$$

Use the first equation to find $c$. Use the third to find $d$. Use those to find $a$ and $b$.

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    $\begingroup$ +1 for converting the polynomial equation into a system of linear equations. You would use the third equation to find $d$. Then the fourth to find $b$, and the second to find $a$. OP, for convenience, here is the equation set arranged as they would be solved via row-reduction. $$\begin{cases} 0a + 0b + c + 0d = -4 \\ -a + 0b + 4c + d = -6 \\ 0a + 0b + 6c + 4d = 0 \\ 0a - b + 0c + 6d = 0 \end{cases}$$ And the solution is: $$a = -4 ~~~ b = 36 ~~~ c = -4 ~~~ d = 6$$ $\endgroup$ – FundThmCalculus Jun 17 '15 at 1:29
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Let $r_1,r_2$ be the roots of $x^2+4x+6$. By Vieta's $r_1+r_2 = -4$ and $r_1r_2 = 6$.

Since $x^2+4x+6$ is a factor of $x^4+ax^2+b$, we have that $r_1,r_2$ are also roots of $x^4+ax^2+b$.

Then, since $x^4+ax^2+b$ is an even polynomial, $-r_1,-r_2$ are also roots of $x^4+ax^2+b$.

It is easy to see that $r_1,r_2,-r_1,-r_2$ are distinct, so we've found all four roots of $x^4+ax^2+b$.

Therefore, $x^4+ax^2+b$ $= (x-r_1)(x-r_2)(x+r_1)(x+r_2)$ $= \left[(x^2-(r_1+r_2)x+r_1r_2\right]\left[(x^2+(r_1+r_2)x+r_1r_2\right]$ $= (x^2+4x+6)(x^2-4x+6)$ $= x^4 - 4x^2 + 36$.

Hence, $a = -4$, $b = 36$, and thus, $a+b = 32$.

Note: I'm not sure how Bezout's Theorem is needed here.

EDIT: Little Bezout's Theorem is also known as the polynomial remainder theorem, which was used above.

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  • $\begingroup$ The quotient could also have been simply presented as $(-x)^2 + 4(-x) + 6$, since the roots are reversed, bypass all of vieta's stuff. $\endgroup$ – DanielV Jun 17 '15 at 5:02
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\begin{align*} x^4+ax^2+b & = x^2\color{blue}{(x^2+4x+6)}-4x\color{blue}{(x^2+4x+6)}+\\ &(a+10)\color{blue}{(x^2+4x+6)}-\color{red}{(4a+16)(x)+(b-6a-60)} \end{align*} For $x^2+4x+6$ to divide the given polynomial.We need $4a+16=0$ and $b-6a-60=0$. Thus $a=-4$ and $b=36$.

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  • $\begingroup$ This doesn't give the correct answer, probably an arithmetic mistake somewhere. $\endgroup$ – DanielV Jun 17 '15 at 0:14
  • $\begingroup$ @DanielV you are right I had made an error in the constant coefficient. I have fixed that. Thanks. $\endgroup$ – Anurag A Jun 17 '15 at 0:34
  • $\begingroup$ I got a positive 36 for b $\endgroup$ – DanielV Jun 17 '15 at 0:34
  • $\begingroup$ @DanielV sorry I did not erase the $-$ from the previous answer. $\endgroup$ – Anurag A Jun 17 '15 at 0:35
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The first solution I came up with involves actually finding the roots of the quadratic, then employing Factor/Remainder theorem, so while it is a tad more tedious, it's also very direct. I will follow with an alternative solution that is more elegant.

By the quadratic formula, roots of $x^2 + 4x + 6$ are the complex conjugate pair $-2 \pm i\sqrt 2$.

By Factor Theorem, those will also be roots of the quartic (biquadratic), giving the linear simultaneous equations:

$(-2 + i\sqrt 2)^4 + a(-2 + i\sqrt 2)^2 + b = 0$

$(-2 - i\sqrt 2)^4 + a(-2 - i\sqrt 2)^2 + b = 0$

Now $(-2 \pm i\sqrt 2)^2 = 2 \mp 4i\sqrt 2$

Squaring again,

$(-2 \pm i\sqrt 2)^4 = -28 \mp 16i \sqrt 2$

which allows you to express the simultaneous equations as:

$(-28 - 16i\sqrt 2) + a(2 - 4i\sqrt 2) + b = 0$

$(-28 + 16i\sqrt 2) + a(2 + 4i\sqrt 2) + b = 0$

by subtraction, we almost immediately have $a(8i\sqrt 2) = -32i\sqrt 2 \implies a = -4$

By adding and back substitution, we get: $-56 + 4(-4) + 2b = 0 \implies b = 36$

And of course, to answer the original question, $a+b= 32$

Alternative solution:

Let the roots of the original quadratic be $x_1, x_2$.

Vieta's formulas give:

$x_1 + x_2 = -4$

$x_1x_2 = 6$

from which we can deduce that:

$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = 4$

and

$x_1^2x_2^2 = 36$

Now, by Factor Theorem, $x_1$ and $x_2$ are also roots of the biquadratic.

If we let $y = x^2$, the biquadratic can be written as:

$y^2 + ay + b = 0$

which will have roots $y_1$ and $y_2$ where $y_1 = x_1^2$ and $y_2 = x_2^2$

Applying Vieta's formula to the quadratic in $y$ allows us to deduce that:

$y_1 + y_2 = -a \implies x_1^2 + x_2^2 = -a$

and

$y_1y_2 = b \implies x_1^2x_2^2 = b$

and by reference to the above results, that allows us to immediately conclude that $a = -4, b = 36$ and $a+b = 32$.

This solution involves Vieta's formulas, and is less tedious than my original method.

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If $x^2 + 4x + 6$ is factor of $x^4 + ax^2 + b$, then it must satify

$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$ where $Q(x)$ is a second degree polynominal. You do not need to find $Q(x)$

$x^2 + 4x + 6=0$

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

$$x^4 + ax^2 + b=(x^2 + 4x + 6)Q(x)$$

$$(-(4x + 6))^2 - a(4x + 6)+ b=0$$

$$16x^2+48x+36 - 4ax -6a+ b=0$$ $$-16(4x + 6))+48x+36 - 4ax -6a+ b=0$$

$$-64x -96+48x+36 - 4ax -6a+ b=0$$

$$-16x -60 - 4ax -6a+ b=0$$

$$-16-4a=0 $$

$$a=-4$$

$$-60 -6a+ b=0$$ $$-60 +24+ b=0$$ $$b=36$$

then $a+b=32$

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