(X+1) (x-1) (x ^2 +1 )
Answers
Answer:
(x+1)(x2−1)
=x(x2)+x(−1)+1(x2)+1(−1)
=x3−x+x2−1
=x3+x2−x−1
Now we have found our new expression. We then multiply this by the third expression to get our final answer;
(x3+x2−x−1)(x−1)
=x(x3)+x3(−1)+x(x2)+x2(−1)+x(−x)+(−x)(−1)+x(−1)+(−1)(−1)
=x4−x3+x3−x2−x2+x−x+1
=x4−2x2+1
We have found the expanded form of our product. You may have realised that this method is very tedious. Luckily, there is a much simpler method, for those who were keen enough to see it.
This method would require us to know a useful algebraic identity. Let a and b be real numbers.
(a+b)(a−b)
=a2−ab+ab−b2
=a2−b2
This is called the “difference of squares”. We have established that (a+b)(a−b)=a2−b2. This will be useful in making our problem much easier.
Now back to the question; look at the first and last expressions, (x+1) and (x−1). This looks like our difference of squares method, where a is x and b is 1. Since Multiplication is commutative, we can simply rearrange the terms and apply the difference of squares method.
(x+1)(x−1)(x2−1)
=[(x+1)(x−1)](x2−1)
=(x2−12)(x2−1)
=(x2−1)(x2−1)
=(x2−1)2
Now we have simplified this expressions, but we still need to expand it. We can use another useful algebraic identity; the “perfect square” method.
(a−b)2
=(a−b)(a−b)
=a2−ab−ab+b2
=a2−2ab+b2
We have established this identity, and now we can see that it is very similar to our expression, where a is x2 and b is 1. We can apply this algebraic identity;
(x2−1)2
=(x2)2−2(x2)(1)+12
=x4−2x2+1
We have arrived at the same final answer as we did using the brute force method.
Therefore, the expanded form of this expression is x4−2x2+1.
Answer:
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