How do I factor a function with with a degree of four?

Question

p(x)= X^4-13X^2+36

Answer 1

put x^2=t the equation reduces to t^2-13t+36
which when factored will be (t-9)(t-4)=(x^2-9)(x^2-4)
=>(x+3)(x-3)(x-2))x+2)

Answer 2

for factorising functions of power more than two there is a methid called division method .
1) u must first write down all the coefficients of the the terms
then from the first tern subtract 0 take some number multiply the remainder to this number this number should be subtracted from the next co efficient and so on. if the difference with the last coefficient is zero then the number u have chosen wil be the root of the function ,to get more roots do the same process for the coefficients obtained while doing the first process
for example the expression u have given can be written as
1 0 -13 0 36. this is because the coeff of the x^3 & x are zero

now take some number say 2apply the process
1)1-0=1(subtracting zero from the 1st coeff)
2)2*1=1(multiply the difference and the chosen number &the ddifference )
3)subtract the number from 0 that is the next coeff we get -2
3)now multiply 2*-2&repeat the process u will notice that the the last difference will be 0
4)that means 2 is a root
5)to find the rest of the roots apply the process to the coeff obtained from the above process i.e 1 -2 -9 18 try them with-2
then go on applying the the process until u get all the roots

remember the main skill in this process is estimation $u get this by a good amount of practice then u can apply this method to any polynomial without any problem

Answer 3

You're in luck because there are no odd powers of x. Treat is as a degree 2 equation in X^2 to get two solutions. Then further solve those second order for solutions in x.

let u=x^2
u^2 - 13u + 36 = (u-9)(u-4)

note that both of these are differences of squares, the easiest to factor.
u-9 = x^2 -9 = (x+3)(x-3)
and
... you can do this one ...

Thus you end up with four real roots of your fourth degree equation. If you have a graphing calculator it's fun to see its shape.

Answer 4

In this case you have a biquadratic equation - simply replace x² with y and factor that: you get y²-13y+36 = (y-4)(y-9). Substitue the x² for y and then factor the factors - i.e. x²-4 = (x+2)(x-2), x²-9=(x+3)(x-3), and so your final factors are (x+2)(x-2)(x+3)(x-3).

Answer 5

p(x)=(x^2-4)(x^2-9)=?

The answer is p(x)=(x-2)(x-3)(x+2)(x+3)

Why?
for x=2 p(2)=0 So 2 is a root
(x-2) is factor.
Divide by (x-2)

x^3 + 2 x^2 -9x-18=0

The value x=3 satisfy the equation.

x^2 +5x+6=0
x=-2 and x=-3 are the roots.

Hint: Learn Horner method

Answer 6

you put x^2 = t then it becomes quadatric in t
t^2- 13t+36
= t^2-9t-4t+36
= t(t-9) -4(t-9)
=(t-9) (t-4)
now (t-9) =(x^2-9) = (x+3)(3-x) using x^2-y^2 = (x+y)(x-y)
(t-4) = x^2- 4 = (x+2)(x-2)
so value = (x+2)(x-2)(x+3)(x-3)

Answer 7

p(x)= (x^2-4) (x^2-9)

so p(x)= (x+2) (x-2) (x+3) (x-3)

if u want p(n), substitute n in ttead of x, that means 'n' is variable value,or n belongs to real numbers.