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Answer :
We are given the division of the polynomial
[tex]$$
3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6
$$[/tex]
by
[tex]$$
x^3.
$$[/tex]
Since the divisor is a monomial, [tex]$x^3$[/tex], we can divide each term of the dividend that has degree at least 3 by [tex]$x^3$[/tex] termâ€byâ€term.
Below is the detailed process:
1. Divide the leading term:
The first term in the dividend is [tex]$3x^5$[/tex]. Divide this by [tex]$x^3$[/tex]:
[tex]$$
\frac{3x^5}{x^3} = 3x^{5-3} = 3x^2.
$$[/tex]
This gives the first term of the quotient.
2. Multiply and subtract:
Multiply the divisor by [tex]$3x^2$[/tex]:
[tex]$$
x^3 \cdot 3x^2 = 3x^5.
$$[/tex]
Subtract this product from the dividend. This subtraction eliminates the [tex]$3x^5$[/tex] term, leaving:
[tex]$$
(3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6) - 3x^5 = 22x^4 + 13x^3 + 39x^2 + 14x + 6.
$$[/tex]
3. Proceed to the next term:
Now, take the new leading term [tex]$22x^4$[/tex] and divide by [tex]$x^3$[/tex]:
[tex]$$
\frac{22x^4}{x^3} = 22x^{4-3} = 22x.
$$[/tex]
This is the next term in the quotient.
4. Multiply and subtract again:
Multiply the divisor by [tex]$22x$[/tex]:
[tex]$$
x^3 \cdot 22x = 22x^4.
$$[/tex]
Subtract this product from the current polynomial:
[tex]$$
(22x^4 + 13x^3 + 39x^2 + 14x + 6) - 22x^4 = 13x^3 + 39x^2 + 14x + 6.
$$[/tex]
5. Divide the next term:
Now, divide the term [tex]$13x^3$[/tex] by [tex]$x^3$[/tex]:
[tex]$$
\frac{13x^3}{x^3} = 13x^{3-3} = 13.
$$[/tex]
Add [tex]$13$[/tex] to the quotient.
6. Multiply and subtract once more:
Multiply the divisor by [tex]$13$[/tex]:
[tex]$$
x^3 \cdot 13 = 13x^3.
$$[/tex]
Subtracting from the current polynomial gives:
[tex]$$
(13x^3 + 39x^2 + 14x + 6) - 13x^3 = 39x^2 + 14x + 6.
$$[/tex]
7. Determine the stopping point:
The remainder is now
[tex]$$
39x^2 + 14x + 6,
$$[/tex]
which has degree [tex]$2$[/tex], lower than the degree [tex]$3$[/tex] of the divisor [tex]$x^3$[/tex]. Thus, the division process stops here.
8. Write the final answer:
The quotient from the division is
[tex]$$
3x^2 + 22x + 13,
$$[/tex]
and the remainder is
[tex]$$
39x^2 + 14x + 6.
$$[/tex]
Thus, we can express the result of the division as:
[tex]$$
\frac{3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6}{x^3} = 3x^2 + 22x + 13 \quad \text{with a remainder of} \quad 39x^2 + 14x + 6.
$$[/tex]
[tex]$$
3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6
$$[/tex]
by
[tex]$$
x^3.
$$[/tex]
Since the divisor is a monomial, [tex]$x^3$[/tex], we can divide each term of the dividend that has degree at least 3 by [tex]$x^3$[/tex] termâ€byâ€term.
Below is the detailed process:
1. Divide the leading term:
The first term in the dividend is [tex]$3x^5$[/tex]. Divide this by [tex]$x^3$[/tex]:
[tex]$$
\frac{3x^5}{x^3} = 3x^{5-3} = 3x^2.
$$[/tex]
This gives the first term of the quotient.
2. Multiply and subtract:
Multiply the divisor by [tex]$3x^2$[/tex]:
[tex]$$
x^3 \cdot 3x^2 = 3x^5.
$$[/tex]
Subtract this product from the dividend. This subtraction eliminates the [tex]$3x^5$[/tex] term, leaving:
[tex]$$
(3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6) - 3x^5 = 22x^4 + 13x^3 + 39x^2 + 14x + 6.
$$[/tex]
3. Proceed to the next term:
Now, take the new leading term [tex]$22x^4$[/tex] and divide by [tex]$x^3$[/tex]:
[tex]$$
\frac{22x^4}{x^3} = 22x^{4-3} = 22x.
$$[/tex]
This is the next term in the quotient.
4. Multiply and subtract again:
Multiply the divisor by [tex]$22x$[/tex]:
[tex]$$
x^3 \cdot 22x = 22x^4.
$$[/tex]
Subtract this product from the current polynomial:
[tex]$$
(22x^4 + 13x^3 + 39x^2 + 14x + 6) - 22x^4 = 13x^3 + 39x^2 + 14x + 6.
$$[/tex]
5. Divide the next term:
Now, divide the term [tex]$13x^3$[/tex] by [tex]$x^3$[/tex]:
[tex]$$
\frac{13x^3}{x^3} = 13x^{3-3} = 13.
$$[/tex]
Add [tex]$13$[/tex] to the quotient.
6. Multiply and subtract once more:
Multiply the divisor by [tex]$13$[/tex]:
[tex]$$
x^3 \cdot 13 = 13x^3.
$$[/tex]
Subtracting from the current polynomial gives:
[tex]$$
(13x^3 + 39x^2 + 14x + 6) - 13x^3 = 39x^2 + 14x + 6.
$$[/tex]
7. Determine the stopping point:
The remainder is now
[tex]$$
39x^2 + 14x + 6,
$$[/tex]
which has degree [tex]$2$[/tex], lower than the degree [tex]$3$[/tex] of the divisor [tex]$x^3$[/tex]. Thus, the division process stops here.
8. Write the final answer:
The quotient from the division is
[tex]$$
3x^2 + 22x + 13,
$$[/tex]
and the remainder is
[tex]$$
39x^2 + 14x + 6.
$$[/tex]
Thus, we can express the result of the division as:
[tex]$$
\frac{3x^5 + 22x^4 + 13x^3 + 39x^2 + 14x + 6}{x^3} = 3x^2 + 22x + 13 \quad \text{with a remainder of} \quad 39x^2 + 14x + 6.
$$[/tex]
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