help me please.:v^4-5v^2+4?

Answer 1

v^4 - 5v^2 + 4 = (v^2 - 4)(v^2 - 1)
=(v+2)(v-2)(v + 1)(v-1)

If this is set equal to zero, then any of v = -2, v = -1, v = 1, or v = 2 satisfies the equation.

Answer 2

The solution to your problem is as follows:

v^4 - 5v^2 + 4

= v^4 - 4v^2 - v^2 + 4

= v^2(v^2 - 4) - 1(v^2 - 1)

= (v^2 - 1) (v^2 - 4)

You can leave the answer here itself or if you want further simplification here it is

= (v + 1)(v - 1) (v + 2)(v - 2)

Answer 3

(v^4 -5v^2 +4) factors to (v^2 -1)(v^2 -4) which are both still factorable as the difference of two squares...(v^2-1) = (v+1)(v-1) and (v^2-4) = (v+2)(v-2)
----for a final answer of (v+1)(v-1)(v+2)(v-2)

Answer 4

v^4 - 5v^2 + 4 = (v^2 - 4)(v^2 - 1) = (v - 2)(v + 2)(v - 1)(v + 1)

Answer 5

The factorization has the form (v^2 + a)(v^2 + b)
where ab=4, and a+b=-5. a and b must be negative, because their sum is negative but their product is positive. Try factors of 4 for a and b. You'll find a=-1 and b=-4 work, so you get

(v^2-4)(v^2-1)

Each factor is the difference of two squares, so you get

(v+2)(v-2)(v+1)(v-1).

Answer 6

This trinomial factors to:

(v^2-4)(v^2-1)

Both of these are difference of squares so you keep going:

(v+2)(v-2)(v+1)(v-1)