Answer :
Sure! Let's divide the polynomial [tex]\(x^4 - 3x^3 - 19x^2 + 87x - 90\)[/tex] by [tex]\((x-3)^2 = x^2 - 6x + 9\)[/tex].
1. Set up the Division:
We want to divide the polynomial by [tex]\(x^2 - 6x + 9\)[/tex].
2. First Division Step:
- Look at the leading terms: [tex]\(x^4\)[/tex] (from the dividend) and [tex]\(x^2\)[/tex] (from the divisor).
- Divide the leading term of the dividend by the leading term of the divisor: [tex]\(\frac{x^4}{x^2} = x^2\)[/tex].
- Multiply the entire divisor by [tex]\(x^2\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times x^2 = x^4 - 6x^3 + 9x^2
\][/tex]
- Subtract this from the original polynomial:
[tex]\[
(x^4 - 3x^3 - 19x^2 + 87x - 90) - (x^4 - 6x^3 + 9x^2) = 3x^3 - 28x^2 + 87x - 90
\][/tex]
3. Second Division Step:
- Divide the new leading term by the leading term of the divisor: [tex]\(\frac{3x^3}{x^2} = 3x\)[/tex].
- Multiply the entire divisor by [tex]\(3x\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times 3x = 3x^3 - 18x^2 + 27x
\][/tex]
- Subtract this from the polynomial:
[tex]\[
(3x^3 - 28x^2 + 87x - 90) - (3x^3 - 18x^2 + 27x) = -10x^2 + 60x - 90
\][/tex]
4. Third Division Step:
- Divide the new leading term by the leading term of the divisor: [tex]\(\frac{-10x^2}{x^2} = -10\)[/tex].
- Multiply the entire divisor by [tex]\(-10\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times -10 = -10x^2 + 60x - 90
\][/tex]
- Subtract this from the polynomial:
[tex]\[
(-10x^2 + 60x - 90) - (-10x^2 + 60x - 90) = 0
\][/tex]
5. Conclusion:
Since the remainder is 0, the division is exact. The quotient is [tex]\(x^2 + 3x - 10\)[/tex].
So, when [tex]\(x^4 - 3x^3 - 19x^2 + 87x - 90\)[/tex] is divided by [tex]\((x-3)^2 = x^2 - 6x + 9\)[/tex], the result is:
[tex]\[
x^2 + 3x - 10
\][/tex]
This means the division is perfect without any remainder.
1. Set up the Division:
We want to divide the polynomial by [tex]\(x^2 - 6x + 9\)[/tex].
2. First Division Step:
- Look at the leading terms: [tex]\(x^4\)[/tex] (from the dividend) and [tex]\(x^2\)[/tex] (from the divisor).
- Divide the leading term of the dividend by the leading term of the divisor: [tex]\(\frac{x^4}{x^2} = x^2\)[/tex].
- Multiply the entire divisor by [tex]\(x^2\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times x^2 = x^4 - 6x^3 + 9x^2
\][/tex]
- Subtract this from the original polynomial:
[tex]\[
(x^4 - 3x^3 - 19x^2 + 87x - 90) - (x^4 - 6x^3 + 9x^2) = 3x^3 - 28x^2 + 87x - 90
\][/tex]
3. Second Division Step:
- Divide the new leading term by the leading term of the divisor: [tex]\(\frac{3x^3}{x^2} = 3x\)[/tex].
- Multiply the entire divisor by [tex]\(3x\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times 3x = 3x^3 - 18x^2 + 27x
\][/tex]
- Subtract this from the polynomial:
[tex]\[
(3x^3 - 28x^2 + 87x - 90) - (3x^3 - 18x^2 + 27x) = -10x^2 + 60x - 90
\][/tex]
4. Third Division Step:
- Divide the new leading term by the leading term of the divisor: [tex]\(\frac{-10x^2}{x^2} = -10\)[/tex].
- Multiply the entire divisor by [tex]\(-10\)[/tex]:
[tex]\[
(x^2 - 6x + 9) \times -10 = -10x^2 + 60x - 90
\][/tex]
- Subtract this from the polynomial:
[tex]\[
(-10x^2 + 60x - 90) - (-10x^2 + 60x - 90) = 0
\][/tex]
5. Conclusion:
Since the remainder is 0, the division is exact. The quotient is [tex]\(x^2 + 3x - 10\)[/tex].
So, when [tex]\(x^4 - 3x^3 - 19x^2 + 87x - 90\)[/tex] is divided by [tex]\((x-3)^2 = x^2 - 6x + 9\)[/tex], the result is:
[tex]\[
x^2 + 3x - 10
\][/tex]
This means the division is perfect without any remainder.
