x x power 4 + 64 factorise it
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x4 + 64 = x4 + 16x2 - 16x2 + 64
The two terms 16x2 and -16x2 appear by considering the square root of x4 and by dividing the original constant by the leading exponent (64 ÷ 4 = 16). Then, you change the position of the negative term to the last position:
x4 + 16x2 - 16x2 + 64 = x4 + 16x2 + 64 - 16x2
The first three terms form a perfect square trinomial, which can be easily factored. The last term can be rewritten considering the square root concept.
x4 + 16x2 + 64 - 16x2
= (x2 + 8)(x2 + 8) - (4x)2
= (x2 + 8)2 - (4x)2
The preceding expression is a difference of two sqaures. Considering the pattern for this type of factorizaion (a2 - b2), aequals x2 + 8 while b = 4x. So, knowing that a2 - b2 = (a + b)(a- b), we have
(x2 + 8)2 - (4x)2 = (x2 + 8 + 4x)(x2 + 8 - 4x),
which gives the solution to the exercise.
The two terms 16x2 and -16x2 appear by considering the square root of x4 and by dividing the original constant by the leading exponent (64 ÷ 4 = 16). Then, you change the position of the negative term to the last position:
x4 + 16x2 - 16x2 + 64 = x4 + 16x2 + 64 - 16x2
The first three terms form a perfect square trinomial, which can be easily factored. The last term can be rewritten considering the square root concept.
x4 + 16x2 + 64 - 16x2
= (x2 + 8)(x2 + 8) - (4x)2
= (x2 + 8)2 - (4x)2
The preceding expression is a difference of two sqaures. Considering the pattern for this type of factorizaion (a2 - b2), aequals x2 + 8 while b = 4x. So, knowing that a2 - b2 = (a + b)(a- b), we have
(x2 + 8)2 - (4x)2 = (x2 + 8 + 4x)(x2 + 8 - 4x),
which gives the solution to the exercise.
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