Finding constants of a quartic equation

[chwala]

Homework Statement

The polynomial $4x^4+Ax^2+11x+b$ where $a$ and $b$ are constants, is denoted by $p(x)$. It is given that $p(x)$ is divisible by $x^2-x+2$. Find the values of $a$ and $b$

Relevant Equations

factor theorem

$4x^4+Ax^2+11x+b$≡$(x^2-x+2)(ax^2+mx+d)$
≡$ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d$

$a=4$
$m-a=0, m=4$
$d-m+2a=A$
$2m-d=11$
$2d=b$

$A=1$, $B=-6$ is that correct?

 

[ehild]

Homework Statement:: The polynomial $4x^4+Ax^2+11x+b$ where $a$ and $b$ are constants, is denoted by $p(x)$. It is given that $p(x)$ is divisible by $x^2-x+2$. Find the values of $a$ and $b$
Relevant Equations:: factor theorem

$4x^4+Ax^2+11x+b$≡$(x^2-x+2)(ax^2+mx+d)$
≡$ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d$

$a=4$
$m-a=0, m=4$
$d-m+2a=A$
$2m-d=11$
$2d=b$

$A=1$, $B=-6$ is that correct?

Yes.

 

[SammyS]

@chwala :
Try to be consistent with labeling of parameters.
In the given quartic polynomial,you use $A$ and $b$. The question then asks for values of $a$ and $b$. Then in the unknown quadratic factor, you use $a$ which is all together different from previous $a$ or $A$, although, by inspection you can determine that this $a=4$ .

In the answer, you have $A$ and $B$.

 

[chwala]

That was a pretty easy question...i guess i was just over thinking on it...

 

[chwala]

@chwala :
Try to be consistent with labeling of parameters.
In the given quartic polynomial,you use $A$ and $b$. The question then asks for values of $a$ and $b$. Then in the unknown quadratic factor, you use $a$ which is all together different from previous $a$ or $A$, although, by inspection you can determine that this $a=4$ .

In the answer, you have $A$ and $B$.

true, i had seen that but i just ignored ...noted

 

[SammyS]

(Using uppercase $A$ and $B$ for the problem), I suppose you could say, if $x^2-x+2$ is a factor of $4x^4+Ax^2-11x+B$, then by inspection, the other factor must be of the form $4x^2+mx+\frac B 2$.

Multiplying the quadratics gives:
$\left(x^2-x+2\right) \left(4x^2+mx+\frac B 2 \right) = 4x^4+(m-4)x^3+\left(8-m+\frac B 2 \right)x^2 +\left(2m-\frac B 2\right)x +B$.

Match coefficients.

etc.

 

[haruspex]

Homework Statement:: The polynomial $4x^4+Ax^2+11x+b$ where $a$ and $b$ are constants, is denoted by $p(x)$. It is given that $p(x)$ is divisible by $x^2-x+2$. Find the values of $a$ and $b$
Relevant Equations:: factor theorem

$4x^4+Ax^2+11x+b$≡$(x^2-x+2)(ax^2+mx+d)$
≡$ax^4+(m-a)x^3+(d-m+2a)x^2+(2m-d)x+2d$

$a=4$
$m-a=0, m=4$
$d-m+2a=A$
$2m-d=11$
$2d=b$

$A=1$, $B=-6$ is that correct?

Not sure it's any easier, but another method is to use $x^2=x-2$ to substitute for $x^2$, first in the $x^4$ term, then in the resulting $x^2$ term.
The resulting linear expression must equal zero.

 

[chwala]

Not sure it's any easier, but another method is to use $x^2=x-2$ to substitute for $x^2$, first in the $x^4$ term, then in the resulting $x^2$ term.
The resulting linear expression must equal zero.

i do not get this approach $x^2=x-2$ how? from where...

 

[haruspex]

i do not get this approach $x^2=x-2$ how? from where...

If p(x) is divisible by $x^2−x+2$ then every solution of $x^2−x+2=0$ makes p(x) zero.

 

[chwala]

aaaaaah...seen that..thanks haruspex