Answer:
![[x + 2][x + 5][x - 1]](https://tex.z-dn.net/?f=%5Bx%20%2B%202%5D%5Bx%20%2B%205%5D%5Bx%20-%201%5D)
Step-by-step explanation:
The Leading Coefficient is 1 and the initial value is 10; they have a common factor of 1, so that automatically gives us our first factor of ![[x - 1]](https://tex.z-dn.net/?f=%5Bx%20-%201%5D)
. Now, since the divisor\factor is in the form of x - c, use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:
1| 1 6 3 −10
↓ 1 7 10
__________
1 7 10 0 → ![{x}^{2} + 7x + 10 >> [x + 2][x + 5]](https://tex.z-dn.net/?f=%7Bx%7D%5E%7B2%7D%20%2B%207x%20%2B%2010%20%3E%3E%20%5Bx%20%2B%202%5D%5Bx%20%2B%205%5D)
You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x³ + 6x² + 3x - 10]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x², the 7x follows right behind it, then 10, giving you the other factor of 
, which can be factored further to ![[x + 2][x + 5]](https://tex.z-dn.net/?f=%5Bx%20%2B%202%5D%5Bx%20%2B%205%5D)
, attaching this to the first factor you started out your work on:
I am joyous to assist you anytime.
The place value of the number 6 is the ten thousands place.
Answer:
Option 4: 4¹¹
Step-by-step explanation:
Looking at the problem, I need to work out 4⁴ squared first, which is the same as 4⁸. Then multiply that by 4³ to get 4¹¹. What I did was simply add 3 + (4 * 2), which is 11.
Answer: There are 1400 different combinations.
Step-by-step explanation:
The conditions are:
we have 4-digits: abcd.
all the digits are different.
a is an odd number, and b is an even number.
Then, for a, we have the options 1, 3, 5, 7 and 9 (so we have 5 options).
for b, we have the options 0, 2, 4, 6 and 8 (so we have 5 options).
for c, we can have odd or even numbers, so we have 8 options ( remember that there where 2 numbers already taken away, this is why we have only 8 options).
for d we have 7 options (because 3 numbers are already taken).
Then the number of combinations is equal to the product of the number of options for each selection:
C = 5*5*8*7 = 1400
