How do you factor x^4-61x^2+900?

Mar 25, 2017

$\left(x + 5\right) \left(x - 5\right) \left(x + 6\right) \left(x - 6\right)$

Explanation:

${x}^{4} - 61 {x}^{2} + 900 = {x}^{4} - \left(25 + 36\right) {x}^{2} + 900$

$\Rightarrow {x}^{4} - 25 {x}^{2} - 36 {x}^{2} + 900$

$\Rightarrow {x}^{2} \left({x}^{2} - 25\right) - 36 \left({x}^{2} - 25\right)$

$\Rightarrow \left({x}^{2} - 25\right) \left({x}^{2} - 36\right)$

$\Rightarrow \left({x}^{2} - {5}^{2}\right) \left({x}^{2} - {6}^{2}\right)$ [here applying ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$]

$\Rightarrow \left(x + 5\right) \left(x - 5\right) \left(x + 6\right) \left(x - 6\right)$

Mar 25, 2017

$= \left(x + 5\right) \left(x - 5\right) \left(x + 6\right) \left(x - 6\right)$

Explanation:

Finding the correct factors for a quadratic trinomial which has a big value for (c) is not as daunting as it might appear at first:

There are a number of clues to look out for.

In $a {x}^{2} + b x + c$ , if $a = 1$, then the difference in size between $b \mathmr{and} c$ is important. If $b$ is quite small it means that the factors you are looking for are close to $\sqrt{c}$

If $b$ is quite large and almost the same size as $c$, then you are working with one very big and one very small factor.

Also remember that an odd number can only come from
ODD + EVEN, so this immediately eliminates the situation of two even factors.

$61$ is quite small compared to $900$

$\sqrt{900} = 30 \mathmr{and} 30 + 30 = 60$

Therefore the required factors are not very far from $30$ and must be odd and even

Consider factors less than $30$ for divisibility into $900$.

29? 28? 27? 26? 25?

Ah, $25$ seems possible.

$900 \div 25 = 36 \text{ and "25 +36 = 61" }$
so these are the factors we want.

${x}^{4} - 61 {x}^{2} + 900$

$= \left({x}^{2} - 25\right) \left({x}^{2} - 36\right)$

$= \left(x + 5\right) \left(x - 5\right) \left(x + 6\right) \left(x - 6\right)$

Jul 8, 2017

A difference approach

$\implies {x}^{4} - 61 {x}^{2} + 900 \text{ "=" } \left(x - 6\right) \left(x + 6\right) \left(x - 5\right) \left(x + 5\right) = 0$

Explanation:

Given:$\text{ } {x}^{4} - 61 {x}^{2} + 900$

Set $\text{ } {x}^{4} - 61 {x}^{2} + 900 = 0$

Set $\beta = {x}^{2}$ then by substitution we have:

${\beta}^{2} - 61 \beta + 900 = 0$

compare to $y = a {x}^{2} + b x + c$

$\beta = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

where ${x}^{2} = \beta \text{; "a=1"; "b=-61"; } c = + 900$

${x}^{2} = \frac{61 \pm \sqrt{{\left(- 61\right)}^{2} - 4 \left(1\right) \left(900\right)}}{2 \left(1\right)}$

${x}^{2} = \frac{61}{2} \pm \frac{11}{2}$

${x}^{2} = 36 \mathmr{and} 25$

$x = \pm 6 \mathmr{and} \pm 5$

$\implies {x}^{4} - 61 {x}^{2} + 900 \text{ "=" } \left(x - 6\right) \left(x + 6\right) \left(x - 5\right) \left(x + 5\right)$ Note about the minimums.

You must not assume that they are centrally located between the points where the plot crosses the x-axis. In fact if you differentiate and set it to 0 you will find that they are slightly off centre.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {x}^{3} - 122 x = 0$

$\implies x = 0 \mathmr{and} \left[\pm \sqrt{30.5} \approx \pm 5.523 \text{ to 3 decimal places}\right]$

Jul 8, 2017

$\left(x - 5\right) \left(x + 5\right) \left(x - 6\right) \left(x + 6\right)$

Explanation:

Call $X = {x}^{2}$, and factor this trinomial:

$f \left(X\right) = {X}^{2} - 61 X + 900$.

Find 2 numbers knowing the sum $\left(b = - 61\right)$,
and the product (c = 900).

To do so, compose factor pairs of $\left(900\right)$

 rarr: ...(-18, -50);(-20, -45);(-25, -36).

This last sum is $\left(- 25 - 36 = - 61 = - b\right) .$

Therefore. the 2 numbers are: $- 25 \mathmr{and} - 36.$

$f \left(X\right) = \left(X - 25\right) \left(X - 36\right) .$

Replace $X \text{ by } {x}^{2}$

$f \left(x\right) = \left({x}^{2} - 25\right) \left({x}^{2} - 36\right) = \left(x - 5\right) \left(x + 5\right) \left(x - 6\right) \left(x + 6\right)$