Which polynomials can't be factored any further? :[?

Question

I'm doing math with a friend and we don't know how to do this...

4x^2+36

16x^2+1

4v^2+12v+9

9y^2-8

Answer 1

1) 4x^2 + 36
4(x^2 + 9)

This cannot be factored any further, without delving into the realm of complex numbers. I'm guessing that, at your current mathematical level, you're dealing with only integer coefficients. In case you're not, this factors as

4(x - 3i)(x + 3i)

2) 16x^2 + 1.

This cannot be factored any further; if it were a minus instead of a plus, we'd have a difference of squares easily. For instance, 16x^2 - 1 = (4x - 1)(4x + 1). But there is no formula for "sums of squares" unless you use complex numbers.

(4x - i)(4x + i)

3) 4v^2 + 12v + 9

This is a perfect square trinomial.
(2v + 3)^2

4) 9y^2 - 8

This cannot be factored without getting into the realm of irrational numbers. If we do get into that realm, this factors as

(3y - sqrt(8) ) (3y + sqrt(8))

Answer 2

Depends on how far you have learnt your in your Maths course.

The first one can be done as 4(x^2+9)

if you know complex numbers, it is 4(x+3i)(x-3i) (i is the square root of -1)

Similarly, 16x^2 + 1 cannot be factored further but with complex numbers it becomes

(4x+i)(4x-i)

4v^2 + 12v + 9 can be factored as (2v+3)^2

9y^2-8 becomes (3y-2sqrt(2))(3+2sqrt(2)) if you have studied surds or irrationals.

Answer 3

yup5 was pretty close:

tho typically "integer coefficients" are usually called for.

note:
16x^2+1 = (4x + i)(4x - i) where [ i^2 = -1]
however, "i" is not an integer:

so
16x^2+1
and
9y^2-8

are your "answers" given the conditions of integer-coefficients

Answer 4

1. 4(x^2 + 9)

2. ???

3. (2v +3)^2

4. (3y + 2sqrt(2) ) (3y - 2 sqrt(2))

THere's your answer