Factor Completely

mathdad

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Factor completely.

4 - 14x^2 - 8x^4

Solution:

What about dividing each term by 2?

2 - 7x^2 - 4x^4

Let u = x^2

Let x^4 = (x^2)^2

2 - 7u - 4u^2

Remaining Steps:

1. Factor by grouping
2. Back-substitute for u

Both steps above should lead me to the answer. Do you agree?
 
You'll have the factor 2 out the front. You can't just divide by 2 and forget about it. In fact, I would factor out -2 rather than 2,
then you can write -2(4x^4 + 7x^2 - 2) and proceed to factorise the trinomial in brackets in your usual way.

When you "back-substitute" for u, look to see if you have the difference of 2 squares which can be factorised further.
 
Factor completely.
4 - 14x^2 - 8x^4
...
What about dividing each term by 2?
We don't want to change the polynomial's value (i.e., reduce it by half). Instead of dividing, what you meant to say is: factor out any common factors, to start.

2*[2 - 7x^2 - 4x^4]

From here on, you're working entirely inside those square brackets, so the complete factorization will be 2 times the remaining factors.


... 1. Factor by grouping
2. Back-substitute for u
... should lead me to the answer ... agree?
No, not in this exercise. You need a third remaining step:

3. Check for factors that can themselves be factored

?
 
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