Factorise x ^ 4 + 1/x ^ 4 - 3

Mathematics

1 answer:

kondaur [170]2 years ago

7 0

Answer:

Step-by-step explanation:

Add 1 and subtract 1, so that the expression won't change.

$x^{4}+\frac{1}{x^{4}}-3=x^{4}+\frac{1}{x^{4}}-3 +1 - 1\\\\=x^{4}+\frac{1}{x^{4}} - 2 + 1\\\\= (x^{2})^{2} + (\frac{1}{x^{2}})^{2}- 2 *x^{2}*\frac{1}{x^{2}} - 1\\\\$

$= (x^{2} -\frac{1}{x^{2}})^{2} - 1\\\\=(x^{2}-\frac{1}{x^{2}})^{2}- 1^{2}$

(a² - b²) = (a+b)(a - b)

Here a = x² - (1/x²) and b = 1

$= [ (x^{2}-\frac{1}{x^{2}})+ 1 ] [ (x^{2} -\frac{1}{x^{2}} - 1 ]$

You might be interested in

between 1995 and 2006, the population of samburg, usa(in thousands) can be modeled by f(x) = 0.35x(squared) - 2.1x+15.8 where x=

Juli2301 [7.4K]

$\bf f(x)=0.35x^2-2.1x+15.8\qquad 
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}$

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.

the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex

well $\bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad a&\quad b &c
\end{array}
\\\\

\\\\
\qquad \textit{vertex of a parabola}\\ \quad \\
\qquad 

\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$