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If x2 and x3 are meant to represent x2 and x3, then x2 times x3 = x5
You find the product of exponent variables by adding the exponents.
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Factor by grouping 3-3x plus x2-x3?
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
What is a grouping factor for polynomial x3 ax 3a 3x2?
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
What is the factor of x3 plus x2 plus 4x plus 4?
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
How many times in a twelve hour period does the sum of the digits on a digital clock equal six and prove it?
notation: natural numbers = 0,1, 2, 3, 4, 5, ....., (some define it without the zero, though) <= means smaller than or equal to, {} is set notation and means a set of numbers : (such that) then some condition. For example {x: x is not a duck} is the set of all things not a duck. Our goal is to prove that there are 21 different times. let x1 = hours, x2 = tens of minutes, x3 = minutes. We are going to prove the statement about the set {x1, x2,x3: 1<=x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. It will be taken by assumption that this set is the set of digital clock combinations that add up to 6. So then, we must prove that there are unique 21 elements in the set {x1 + x2 + x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. {x1 , x2 , x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6} = {x1 , x2 , x3 : 1<= x1 <= 6, 0<= x2<=5, 0<=x3 <= 5, x1 + x2 + x3 = 6} because x3<=6, and because if x1 >=1, then x2 + x3 <=5, and x3, x2 >= 0 , so surely x3, x2 <= x5. Either x1 = 1, 2, 3, 4, 5, or 6. Next, x1 + x2 + x3 = 6, so x2 + x3 = 6 - x1. There are n+1 natural numbers between 0 and n (I'm being lazy and not proving this, but the proof would be so much longer if I proved it), and since 0 <= x2 <= 5 <= 6-x1, there are at most 6-x1+1 values of x2 for each value of x1. When x1 = 1, there are a maximum of 6, when x1 = 2, there are 6-2+1 = 5, when x1 = 3, there are 6-3+1 = 4, when x1 = 3, there are 3, then 2, and then 1. Summing this up gives us a maximum of 21. So it is at most 21 and at least 21, so exactly 21.
What is x3 plus 4x2-3x-12 divided by x2-3?
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
What times x2-x3?
-1
What is x3-x2-x minus 2 factored?
x3-x2
How do you factor x3?
X3 X(X2) X2(X) and, X * X * X
Factor by grouping 3-3x plus x2-x3?
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
How do you factor x3 plus x2-3x-3?
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
What is x3-x2 plus 2x factored?
x3 - x2 + 2x = x*(x2 - x + 2) which cannot be factored further.
What is x3 - x2 plus x - 2 factored?
x3 - x2 + x - 2 has no rational factors.
What is a grouping factor for polynomial x3 ax 3a 3x2?
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
How do you factor x3-x2?
Answer: x (x2-x)
What is the factor of x3 plus x2 plus 4x plus 4?
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
X3 plus x2 minus 6x plus 4?
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
What is X squared times X?
x cubed, or x^3 x(x2)= x3
