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

4 Answers
Mar 25, 2017

Answer:

#(x+5)(x-5)(x+6)(x-6)#

Explanation:

#x^4-61x^2+900 = x^4-(25+36)x^2+900#

#rArrx^4-25x^2-36x^2+900#

#rArr x^2(x^2-25)-36(x^2-25)#

#rArr(x^2-25)(x^2-36)#

#rArr (x^2-5^2)(x^2-6^2)# [here applying #a^2-b^2=(a-b)(a+b)#]

#rArr (x+5)(x-5)(x+6)(x-6)#

Answer:

#=(x+5)(x-5)(x+6)(x-6)#

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 #ax^2 + bx +c# , if #a=1#, then the difference in size between #b and c# is important. If #b# is quite small it means that the factors you are looking for are close to #sqrtc#

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#

#sqrt900 = 30 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" and "25 +36 = 61" "#
so these are the factors we want.

#x^4 -61x^2 +900#

#=(x^2 -25)(x^2-36)#

#=(x+5)(x-5)(x+6)(x-6)#

Jul 8, 2017

Answer:

A difference approach

#=>x^4-61x^2+900 " "=" "(x-6)(x+6)(x-5)(x+5)=0#

Explanation:

Given:#" "x^4-61x^2+900#

Set #" "x^4-61x^2+900=0#

Set #beta =x^2# then by substitution we have:

#beta^2-61beta+900=0#

compare to #y=ax^2+bx+c#

#beta=(-b+-sqrt(b^2-4ac))/(2a)#

where #x^2=beta"; "a=1"; "b=-61"; "c=+900#

#x^2=(61+-sqrt((-61)^2-4(1)(900)))/(2(1))#

#x^2=61/2+-11/2#

#x^2=36 and 25#

#x=+-6 and +-5#

#=>x^4-61x^2+900 " "=" "(x-6)(x+6)(x-5)(x+5)#

Tony B

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.

#dy/dx=4x^3-122x=0#

#=>x=0 and [+-sqrt(30.5)~~+-5.523" to 3 decimal places"] #

Jul 8, 2017

Answer:

#(x - 5)(x + 5)(x - 6)(x + 6)#

Explanation:

Call #X = x^2#, and factor this trinomial:

#f(X) = X^2 - 61X + 900#.

Find 2 numbers knowing the sum #(b = - 61)#,
and the product (#c = 900)#.

To do so, compose factor pairs of #(900)#

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

This last sum is #(-25 - 36 = - 61 = - b). #

Therefore. the 2 numbers are: #- 25 and - 36.#

#f(X) = (X - 25)(X - 36).#

Replace #X " by " x^2#

#f(x) = (x^2 - 25)(x^2 - 36) = (x - 5)(x + 5)(x - 6)(x + 6)#