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
. 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
, attaching this to the first factor you started out your work on:
![[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)
I am joyous to assist you anytime.
1) substitute 10x in for the equation.
So the vs eliminate each other which leaves the ws and the 1,7. Then you add the 2 ws to get 2w and then add the numbers to get 2w=8, then divide 8 by 2 and you get w=4
The two rational expressions will be; (x + 2)/(x² - 36) and 1/(x² + 6x)
<h3>How to simplify Quadratic Expressions?</h3>
We want to determine the two rational expressions whose difference completes the equation.
The two rational expressions will be;
(x + 2)/(x² - 36) and 1/(x² + 6x)
Now, this can be proved as follows;
Step 2 [(x + 2)/(x² - 36)] - [1/(x² + 6)]
= [(x + 2)/(x + 6)(x - 6)] - [1/(x(x + 6)]
Step 3; By subtracting, we have;
[x(x + 2) - (x - 6)]/[x(x + 6)(x - 6)]
Step 4; By further simplification of step 3, we have;
[x² + x + 6]/[x(x-6)(x + 6)]
Read more about Quadratic Expressions at; brainly.com/question/1214333
#SPJ1
Answer:
median then mean then range