solving the value for (x)

**srirahulan** · June 28th 2012, 02:37 AM

$6x^4+7x^3-36x^2-7x+6=0$ how can i solve this equation and find out the values for (x).

**a tutor** · June 28th 2012, 02:49 AM

Factorise.

**biffboy** · June 28th 2012, 04:06 AM

Originally Posted by srirahulan

$6x^4+7x^3-36x^2-7x+6=0$ how can i solve this equation and find out the values for (x).

Expression =0 when x=2 so (x-2) is a factor. Do long division to investigate other factors.

**srirahulan** · June 28th 2012, 07:43 AM

My question is how do you get (x-2) is a factor.i can't understand about your process.

**biffboy** · June 28th 2012, 08:06 AM

It is a general rule that if a polynomial is zero when x=a then (x-a) is a factor of the polynomial

**Reckoner** · June 28th 2012, 08:19 AM

Originally Posted by srirahulan

My question is how do you get (x-2) is a factor.i can't understand about your process.

This can be found with trial and error. We know, from the rational root theorem, that every rational root of a polynomial with integer coefficients must be of the form $\textstyle\frac pq,$ where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. The constant term in this polynomial is 6 which has factors 1, 2, 3, and 6, and the leading coefficient is also 6. That means the possible rational roots are

$\pm1, \pm2, \pm3, \pm6, \pm\frac12, \pm\frac13, \pm\frac16, \pm\frac23,\ \mathrm{and}\ \pm\frac32.$

By testing each of these with synthetic division, we find that 2 is a root of the polynomial and hence $(x - 2)$ is a factor (we can also determine that $\textstyle-3,-\frac12,$ and $\textstyle\frac13$ are roots).

**srirahulan** · June 28th 2012, 04:16 PM

Great Thanks.I can understand about it.

**thevinh** · June 29th 2012, 10:24 AM

$\text{This is a palindromic equation of even degree, meaning it has the form:}$

$a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\dots+a_2x^2+a_1x+a_0=0.$

$\text{Realizing that}\mspace{7mu} x=0 \mspace{7mu} \text{is not the solution of the equation, thus dividing both side by}\mspace{7mu} x^2\mspace{7mu} \text{we would get:}$

$6x^2+7x-36-\frac{7}{x}+\frac{6}{x^2}=0\, \Leftrightarrow\, 6\left(x^2+\frac{1}{x^2}\right)+7\left(x-\frac{1}{x}\right)-36=0$
$6\left(x^2+\frac{1}{x^2}-2+2\right)+7\left(x-\frac{1}{x}\right)-36=0$
$\Leftrightarrow\, 6\left(x-\frac{1}{x}\right)^2+7\left(x-\frac{1}{x}\right)-24=0$
$\text{Let} \mspace{7mu} u=x-\frac{1}{x}\mspace{7mu} \text{we get a quadratic equation in terms of \emph{u}}.$
$6u^2+7u-24=0\, \Leftrightarrow\, (3u+8)(2u-3)=0\, \Leftrightarrow\, u=-\frac{8}{3} \mspace{7mu} \lor \mspace{7mu} u=\frac{3}{2}$
$\text{When}\mspace{7mu} u=-\frac{8}{3}\, \Rightarrow\, x-\frac{1}{x}=-\frac{8}{3}$

$\text{You can finish the rest starting from here. Have fun! This is the general method to solve a palindromic equation of even degree.}$

$\text{Some equation might not have rational roots, therefore you would have to resort to this method.}$

**HallsofIvy** · June 29th 2012, 11:35 AM

By the "rational root theorem" we can see that any possible rational number roots must be fractions having numerators and denominators factors of 6. A quick search for integer roots shows that 2 and -3 both satisfy the equation and so x- 2 and x+ 3 are factors. Dividing by the those leaves $6x^2+ x- 1= (3x- 1)(2x+ 1)$ so that x= 1/3 and x= -1/2 are the other two roots.

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