IDNLearn.com offers a collaborative platform for sharing and gaining knowledge. Find in-depth and trustworthy answers to all your questions from our experienced community members.
Sagot :
To solve the polynomial division [tex]\( \frac{x^5 - 4x^3 + x + 1}{x^2 - 1} \)[/tex], we'll use long division of polynomials. Let's go through this step-by-step:
1. Set up the division:
[tex]\[ \frac{x^5 - 4x^3 + x + 1}{x^2 - 1} \][/tex]
2. Divide the leading term of the numerator by the leading term of the denominator:
[tex]\[ \frac{x^5}{x^2} = x^3 \][/tex]
So, our first term in the quotient is [tex]\( x^3 \)[/tex].
3. Multiply the entire denominator by [tex]\( x^3 \)[/tex]:
[tex]\[ x^3 \cdot (x^2 - 1) = x^5 - x^3 \][/tex]
4. Subtract this result from the original polynomial:
[tex]\[ (x^5 - 4x^3 + x + 1) - (x^5 - x^3) = -3x^3 + x + 1 \][/tex]
Now, our new polynomial is [tex]\( -3x^3 + x + 1 \)[/tex].
5. Repeat the process with the new polynomial. Divide the new leading term by the leading term of the denominator:
[tex]\[ \frac{-3x^3}{x^2} = -3x \][/tex]
So, our next term in the quotient is [tex]\( -3x \)[/tex].
6. Multiply the entire denominator by [tex]\( -3x \)[/tex]:
[tex]\[ -3x \cdot (x^2 - 1) = -3x^3 + 3x \][/tex]
7. Subtract this result from the new polynomial:
[tex]\[ (-3x^3 + x + 1) - (-3x^3 + 3x) = -2x + 1 \][/tex]
Now, our new polynomial is [tex]\( -2x + 1 \)[/tex].
8. Divide the leading term of the new polynomial by the leading term of the denominator:
[tex]\[ \frac{-2x}{x^2} = -2 \cdot \frac{1}{x} \][/tex]
However, since [tex]\(\frac{-2x}{x^2}\)[/tex] does not provide a polynomial term without a negative power of [tex]\(x\)[/tex], our polynomial division stops here.
9. The quotient and the remainder:
- Quotient: By combining the terms we have calculated, the quotient is [tex]\( x^3 - 3x \)[/tex].
- Remainder: The remaining part of the polynomial that we could not divide further is [tex]\( -2x + 1 \)[/tex].
Thus, the polynomial division gives us the final result:
[tex]\[ \frac{x^5 - 4x^3 + x + 1}{x^2 - 1} = x^3 - 3x \quad \text{with a remainder of} \quad -2x + 1 \][/tex]
So we can write:
[tex]\[ x^5 - 4x^3 + x + 1 = (x^2 - 1)(x^3 - 3x) + (-2x + 1) \][/tex]
1. Set up the division:
[tex]\[ \frac{x^5 - 4x^3 + x + 1}{x^2 - 1} \][/tex]
2. Divide the leading term of the numerator by the leading term of the denominator:
[tex]\[ \frac{x^5}{x^2} = x^3 \][/tex]
So, our first term in the quotient is [tex]\( x^3 \)[/tex].
3. Multiply the entire denominator by [tex]\( x^3 \)[/tex]:
[tex]\[ x^3 \cdot (x^2 - 1) = x^5 - x^3 \][/tex]
4. Subtract this result from the original polynomial:
[tex]\[ (x^5 - 4x^3 + x + 1) - (x^5 - x^3) = -3x^3 + x + 1 \][/tex]
Now, our new polynomial is [tex]\( -3x^3 + x + 1 \)[/tex].
5. Repeat the process with the new polynomial. Divide the new leading term by the leading term of the denominator:
[tex]\[ \frac{-3x^3}{x^2} = -3x \][/tex]
So, our next term in the quotient is [tex]\( -3x \)[/tex].
6. Multiply the entire denominator by [tex]\( -3x \)[/tex]:
[tex]\[ -3x \cdot (x^2 - 1) = -3x^3 + 3x \][/tex]
7. Subtract this result from the new polynomial:
[tex]\[ (-3x^3 + x + 1) - (-3x^3 + 3x) = -2x + 1 \][/tex]
Now, our new polynomial is [tex]\( -2x + 1 \)[/tex].
8. Divide the leading term of the new polynomial by the leading term of the denominator:
[tex]\[ \frac{-2x}{x^2} = -2 \cdot \frac{1}{x} \][/tex]
However, since [tex]\(\frac{-2x}{x^2}\)[/tex] does not provide a polynomial term without a negative power of [tex]\(x\)[/tex], our polynomial division stops here.
9. The quotient and the remainder:
- Quotient: By combining the terms we have calculated, the quotient is [tex]\( x^3 - 3x \)[/tex].
- Remainder: The remaining part of the polynomial that we could not divide further is [tex]\( -2x + 1 \)[/tex].
Thus, the polynomial division gives us the final result:
[tex]\[ \frac{x^5 - 4x^3 + x + 1}{x^2 - 1} = x^3 - 3x \quad \text{with a remainder of} \quad -2x + 1 \][/tex]
So we can write:
[tex]\[ x^5 - 4x^3 + x + 1 = (x^2 - 1)(x^3 - 3x) + (-2x + 1) \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.