Can someone give me step by step instructions on how to solve this problem?

Question

Can someone give me step by step instructions on how to solve this problem? Please do not just give the answer. I need to know the method of solving a problem like this. Thanks!

Factor: b^8 - 100

I need the steps to be typed as if I did not known anything about math.

Answer 1

This is a difference of two squares.

Normal factorisation is (sum of roots)(difference of roots)

Or, if you prefer symbols:

a^2 - b^2 = (a+b)(a-b).

Here, b^8 = square of b^4
100 is square of 10

b^8 - 100 = (b^4 + 10)(b^4 - 10)

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the easy stuff ends here. If this is at the introduction level, then that is enough. If you are into higher level maths, then:

You could continue this, as long as you have "squares" and depending what you allow yourself to consider a "number"

b^4 - 10 = (b^2 + SQRT(10))(b^2 - SQRT(10))

b^2 - SQRT(10) = (b + root4(10))(b-root4(10))
where root4(10) means the 4th root of 10 (approximately 1.77828)

b^8 - 100 =
(b^4 + 10)(b^2 + SQRT(10))(b + root4(10))(b - root4(10))

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If you use complex numbers (where you are allowed to find square roots of negative numbers), you could continue with the other factor

b^4 + 10 = b^4 - (-10) = (b^2 + SQRT(-10))(b^2 - SQRT(-10))
(where the square root of -1 is normally indicated by the letter i ).

b^8 - 100 =
{b + [(i-1)*root4(5/2)]}*
{b + [(i-1)*root4(5/2)]}}*
{b+[(i+1)*root4(5/2)]}*
{b-[(i+1)*root4(5/2)]}*
{b + i*root4(10)}*{b - i*root4(10)}*
{b + root4(10)}*{b - root4(10)}

This would give you all eight complex roots of b^8-100, including the two that are real numbers.

Answer 2

Think of it as a difference of squares:

b^8 - 100

= (b^4)^2 - 10^2

= (b^4 + 10)(b^4 - 10)

In general, a difference of squares factors as follows:

a^2 - b^2 = (a + b)(a - b)

Answer 3

This factors as the difference of squares.
(a² - b²) = (a + b)*(a - b) so
b^8 - 100 = (b^4 + 10)*(b^4 - 10) the 2'nd term can again be factored as
b^4 - 10 = (b² + √10)*(b² - √10) and, again, the 2'nd term can be further factored into
b² - √10 = (b + √√10)*(b - √√10) so the final factorization is
(b^4 + 10)*(b² + √10)*(b + √√10)*(b - √√10)

Doug

Answer 4

This is typical of the method called difference of 2 squares

c² - d² = (c - d)(c + d)

so b^8 - 100 = (b^4 - 10)(b^4 + 10)

...................... = (b² - √10)(b² + √10)(b^4 + 10)

...................... = (b - 10^(1/4))(b + 10^(1/4)(b² + √10)(b^4 + 10)

Answer 5

First, multiply the -6 and you will do away with the parentheses -12b + -6 + 13b -7 = 0 Regroup and upload what you are able to (-12b + 13b) + (-6-7)=0 b-13=0 to do away with the 13 b-6+13=0+13 b=13

Answer 6

This is the difference of two squares.

b^8 - 100 = (b^4 - 10)(b^4 + 10)

If the power of b is even then you can always do this.

b^(2p) - r = (b^p + √r)(b^p - √r)

Answer 7

we use x^2-y^2=(x+y)(x-y)...........................(I)

b^8 - 100
=(b^4)^2-10^2
=(b^4+10)(b^4-10) [by (I)]