I need help urgently!

HelpMehhhhh

New member
Joined
Oct 22, 2020
Messages
24
As the title suggests I need help understanding and doing a solving question which has thoroughly confused me for a few days, I'd appreciate anything you figure out about it. "Determine b and Q(x) if x^2 + bx + 3 = (x + 4) ∙ Q(x) − 17", you need find bot Q and b but I don't understand how
 
Why is b = 0 or Q = 0? Neither of those follows from x = 0.

x^2 + bx + 3 = (x + 4) ∙ Q(x) − 17

x = 0

3 = 4 ∙ Q(0) − 17 ==> Q(0) = (3 + 17)/4 = 5 and we know nothing about b.

Heard of the Remainder Theorem? You'll need it.
 
As the title suggests I need help understanding and doing a solving question which has thoroughly confused me for a few days, I'd appreciate anything you figure out about it. "Determine b and Q(x) if x^2 + bx + 3 = (x + 4) ∙ Q(x) − 17", you need find bot Q and b but I don't understand how
x^2 +bx +3 is a quadratic. Since (x+4)Q(x) - 7 equals a quadratic then it too must be a quadratic. Since 17 is just a constant it must be that (x+4)Q(x) has the x^2 term. That means that Q(x) must be in the x+s. Now simplify (x+4)(x+s)-17 and set it equal to x^2 + bx + 3
 
Why is b = 0 or Q = 0? Neither of those follows from x = 0.

x^2 + bx + 3 = (x + 4) ∙ Q(x) − 17

x = 0

3 = 4 ∙ Q(0) − 17 ==> Q(0) = (3 + 17)/4 = 5 and we know nothing about b.

Heard of the Remainder Theorem? You'll need it.

Thank you for explaining that, but for the remainer theorm what root would I use? would 5 be a root?
 
Consider long division:

[math]\dfrac{x^{2}+bx + 3}{x+4} = Q(x) - \dfrac{17}{x+4}[/math]
 
Thank you for explaining that, but for the remainer theorm what root would I use? would 5 be a root?
Let's see if 5 is a root.

x^2 + bx + 3 = (x + 4) ∙ Q(x) − 17
25 + 5b + 3 =9Q(5)-17 = 0? Why should it equal to 0??
 
Top