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(B^-1)B = 1
B^(-1xB) = 1/(B^B)
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∙ 13y ago
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How do you completely factorise x to the fourth power minus x to the third power minus x minus 1?
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.
How do you factor completely x to the fourth power minus x squared minus 2?
(x2 + 1)(x2 - 2)
What is x times a minus b equals m times x minus c for x?
x(-b)=m(x-c)
How minus times minus is plus?
Let y= ab+(- a)(b) +(-a)(-b) factor out -a y= ab+(-a){b+(-b)} y=ab+(-a)(0) y =ab -------------------(1) now factor out b y= b{a+(-a)}+(-a)(-b) y= b(0) +(-a)(-b) y= (-a)(-b)-----------------(2) equate (1) and (2) (-a)(-b)=ab minus x minus = positive
How do you factor x to the third power minus one?
(x - 1)(x^2 + x + 1)
What is the factor of x to the third power minus x?
x(x + 1)(x - 1)
How do you factor x to the second power minus x?
x(x - 1)
How do you completely factorise x to the fourth power minus x to the third power minus x minus 1?
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.
What are factors of x2-1?
if you mean x to the second power minus 1 then the answer is (x+1)(x-1)
How do you factor x to the power of b plus 1 minus x to the power of b?
It's hard to write this on a typewriter keyboard. Do you mean xb + (1-x)b , or xb + 1 - xb , or maybe even x(b+1) - xb ? The first cannot be factored. The second equals 1, so it is already prime (cannot be factored). The third could be written as xb(x) - xb(1), so there is a common factor of xb, and the expression can be factored as xb(x-1). This is prime.
What is the answer to factor the following polynomial x to the second power minus one?
(x + 1)(x - 1)
What is the integral of 1 over c to the power of 1 minus x?
use this strategy: integral of (b^x) dx = (b^x)/ln(b) + K [K is integration constant, b is not a variable]rewrite (1/c)^(1-x) = ((1/c)^1)*((1/c)^(-x)) = (1/c)*(c^x). (1/c) is a constant, so bring outside the integral, then let b = c in the formula above, and you have (1/c)*(c^x)/ln(c) + K
How do you factor completely x to the fourth power minus x squared minus 2?
(x2 + 1)(x2 - 2)
What is 3 to the power of minus 2 X 3 to the power of minus 3?
3-2 x 3-3 = 3-5 or 1/243
1 divided by x squared?
That's the same as x to the power minus two.
What is x times a minus b equals m times x minus c for x?
x(-b)=m(x-c)
Why everything to power of zero is equal to one?
x to the power a divided by x to the power b is x to the power (a - b) When a and b are equal then this is x to the power a divided by x to the power a, ie 1. x to the power (a - a) = x to the power zero. Things which are equal to the same thing are equal to each other, so 1 = x to the power zero.
