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Let [tex]$a(x)=5x^2-6x+10x-2$[/tex], and [tex]$b(x)=15x^3+2x$[/tex].

When dividing [tex][tex]$a$[/tex][/tex] by [tex]$b$[/tex], we can find the unique quotient polynomial [tex]$q$[/tex] and remainder polynomial [tex][tex]$r$[/tex][/tex] that satisfy the following equation:

[tex]\frac{a(x)}{b(x)}=q(x)+\frac{r(x)}{b(x)},[/tex]

where the degree of [tex]$r(x)$[/tex] is less than the degree of [tex]$b(x)$[/tex].

What is the quotient, [tex][tex]$q(x)$[/tex][/tex]?
[tex]q(x)=[/tex] [tex]$\square$[/tex]

What is the remainder, [tex]$r(x)$[/tex]?
[tex]r(x)=[/tex] [tex]$\square$[/tex]



Answer :

GinnyAnswer
Let's solve the given problem step-by-step.

1. Combine like terms in [tex]\( a(x) \)[/tex]:
[tex]\[ a(x) = 5x^2 - 6x + 10x - 2 \][/tex]
Combine the [tex]\( x \)[/tex] terms:
[tex]\[ a(x) = 5x^2 + 4x - 2 \][/tex]

2. Identify the degrees of [tex]\( a(x) \)[/tex] and [tex]\( b(x) \)[/tex]:
[tex]\[ \deg(a(x)) = 2 \quad \text{(since the highest power of \( x \) in \( a(x) \) is 2)} \][/tex]
[tex]\[ \deg(b(x)) = 3 \quad \text{(since the highest power of \( x \) in \( b(x) \) is 3)} \][/tex]

3. Perform polynomial division:
Since the degree of [tex]\( a(x) \)[/tex] is less than the degree of [tex]\( b(x) \)[/tex], the quotient [tex]\( q(x) \)[/tex] is 0.

4. Determine the remainder:
The remainder [tex]\( r(x) \)[/tex] is simply the polynomial [tex]\( a(x) \)[/tex] itself, because [tex]\( q(x) \cdot b(x) \)[/tex] would be 0 and the remaining [tex]\( a(x) \)[/tex] is unchanged.

Thus:
- The quotient [tex]\( q(x) \)[/tex]:
[tex]\[ q(x) = 0 \][/tex]

- The remainder [tex]\( r(x) \)[/tex]:
[tex]\[ r(x) = 5x^2 + 4x - 2 \][/tex]

Therefore, the final answer is:
[tex]\[ q(x) = 0 \][/tex]
[tex]\[ r(x) = 5x^2 + 4x - 2 \][/tex]

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