factorise the expressions x square minus 1 upon x square
Answers
Answer:
a "difference" means a "subtraction". So a difference of squares is something that looks like x2 – 4. That's because 4 = 22, so we really have x2 – 22, which is a difference of squares.
Step-by-step explanation:
To factor this, I'll start by writing my parentheses, in the same way as usual for factoring:
x2 – 4 = (x )(x )
For this quadratic factorization, I need factors of –4 that add up to zero, so I'll use –2 and +2:
x2 – 4 = (x – 2)(x + 2)
Note that we had x2 – 22, and ended up with (x – 2)(x + 2). Differences of squares (being something squared minus something else squared) always work this way:
For a2 – b2, I start by doing the parentheses:
( )( )
Then I put the first squared thing in front:
(a )(a )
...and I put the second squared thing in back:
(a b)(a b)
...and then I alternate the signs in the middles:
(a – b)(a + b)
By the way, no, the order of the factors doesn't matter. Since multiplication is commutative (that is, since you can move the factors around without changing the value of the product), the difference of squares can also be stated as:
(a + b)(a – b)
Don't get hung up on the order of the factors. Either way is fine.
Here are examples of some typical homework problems:
Factor x2 – 16
This quadratic can be restated as x2 – 42, which is a difference of squares. Applying the formula, I get:
x2 – 16 = x2 – 42
= (x – 4)(x + 4)
Factor 4x2 – 25
This quadratic is (2x)2 – 52 so, applying the formula, I get:
4x2 – 25 = (2x)2 – 52
= (2x – 5)(2x + 5)
Factor 9x6 – y8
This can be restated as (3x3)2 – (y4)2, so I get:
9x6 – y8 = (3x3)2 – (y4)2
= (3x3 – y4)(3x3 + y4)
Factor x4 – 1
This is (x2)2 – 12 so, applying the formula, I get:
x4 – 1 = (x2)2 – 12
= (x2 – 1)(x2 + 1)
Note that I'm not done yet, because one of the factors I got — namely, the x2 – 1 factor — is itself a difference of squares, so I need to apply the formula again to get the fully-factored form. Since x2 – 1 = x2 – 12 = (x – 1)(x + 1), then:
x4 – 1 = (x2)2 – 12
= (x2 – 1)(x2 + 1)
= ((x)2 – (1)2)(x2 + 1)
= (x – 1)(x + 1)(x2 + 1)
The answer to this last exercise depended on the fact that 1, to any power at all, is still just 1.