Polinomial (Bagian 4) - Teorema Sisa dan Teorema Faktor

Main ideas / concepts covered

The video (Part 4 of a polynomial series) focuses on two main tools:

1. Remainder Theorem (three theorems) for finding the remainder of polynomial division without long division.
2. Factor Theorem for determining whether a polynomial has a given polynomial factor, using the condition remainder = 0.

Background idea used throughout:

• When a polynomial (f(x)) is divided by a divisor, it can be written as [ f(x) = (\text{divisor})\cdot(\text{quotient}) + (\text{remainder}). ]

• If you know the divisor’s “zero/solution” value, you can substitute into (f(x)) to get the remainder quickly.

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Methodology / steps (detailed)

A) Remainder Theorem #1

Statement (as conveyed):

• The remainder of dividing (f(x)) by (x+k) is: [ \text{remainder} = f(-k). ]

• Key method: find the zero-maker of the divisor (x+k=0), then substitute that value into (f(x)).

Steps:

1. Identify the divisor (x+k).
2. Solve (x+k=0) → zero-maker is (x=-k).
3. Compute the remainder as (f(-k)) by substituting (x=-k) into the polynomial.
4. The resulting number is the remainder.

Example shown:

• Find the remainder of [ x^4 - 2x^3 + 4x^2 - 5 \quad \text{divided by} \quad x-1. ]

• Zero-maker: (x-1=0 \Rightarrow x=1).

• Substitute (x=1):
  • remainder (= f(1) = -2) (as computed in the subtitles).

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B) Remainder Theorem #2

Statement (as conveyed):

• The remainder of dividing (f(x)) by (x-b) is: [ \text{remainder} = f(b). ]

• Key method: find the zero-maker of the divisor, then substitute into (f(x)).

Steps:

1. Identify the divisor in the form (x-b).
2. Solve (x-b=0) → zero-maker is (x=b).
3. Compute remainder as (f(b)) by substitution.

Example shown:

• Find the remainder of [ x^3 - 2x^2 + 3 \quad \text{divided by} \quad 2x+3. ]

• Convert/zero-maker: [ 2x+3=0 \Rightarrow x=-\frac{3}{2}. ]

• Substitute (x=-\frac{3}{2}) into the polynomial.

• The subtitles conclude the remainder is (-\frac{39}{8}) (noting the arithmetic is messy in the subtitles, but the final stated remainder is (-39/8)).

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C) Remainder Theorem #3 (Quadratic divisor)

Statement (as conveyed):

• When dividing (f(x)) by a quadratic of the form [ (x-a)(x-b) \quad \text{(or)} \quad x^2 - (\text{something})x + (\text{something}), ] the remainder is a linear polynomial: [ \text{remainder} = Px + Q. ]

• You determine (P) and (Q) using known remainder values at special inputs.

Steps (as demonstrated in the example):

1. Let the remainder be: [ Px+Q. ]

2. Use the fact that:

   • If (x=a) makes the quadratic zero, then the remainder at (x=a) equals the remainder from dividing by the corresponding linear factor.
   • The subtitles use given remainders at:
     • (x=-2) (from divisor (x+2))
     • (x=4) (from divisor (x-4))

3. Substitute those (x)-values into (Px+Q) to create equations:

   • If the remainder when divided by (x+2) is (8), then: [ P(-2)+Q=8. ]

   • If the remainder when divided by (x-4) is (-4), then: [ P(4)+Q=-4. ]

4. Solve the system for (P) and (Q) (the subtitles show elimination).

5. Write the remainder polynomial (Px+Q).

Example shown:

• Given:
  • remainder of (f(x)) when divided by (x+2) is (8)
  • remainder of (f(x)) when divided by (x-4) is (-4)

• Find the remainder when dividing by: [ x^2 - 2x - 8. ]

• Factor the quadratic: [ x^2 - 2x - 8 = (x+2)(x-4). ]

• Conclude remainder: [ Px+Q = -2x+4, ] so the remainder stated is (-2x+4).

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D) Factor Theorem

Definition / idea

• A polynomial (g(x)) is a factor of (f(x)) iff the remainder upon dividing (f(x)) by (g(x)) is 0.
• In particular, for a linear factor (x-k): [ (x-k) \text{ is a factor of } f(x) \iff f(k)=0. ]

Method / steps (as demonstrated)

1. If (x-k) is a candidate factor, compute (f(k)).
2. If (f(k)=0), then (x-k) is confirmed as a factor.
3. To find other factors:
   • Use polynomial division (the subtitles mention synthetic division / Horner method).
   • If the division remainder is (0), the divisor is a factor.
   • The quotient becomes the remaining factor.

Example shown:

• Polynomial: [ x^3 + x^2 + x - 3. ]

• Given one factor: [ x-1. ]

• Check using the remainder theorem: [ f(1)=1^3+1^2+1-3=0, ] so (x-1) is a factor.

• Divide (x^3 + x^2 + x - 3) by (x-1) using Horner method.

• Quotient stated: [ x^2 + 2x + 3. ]

• Therefore: [ (x-1)(x^2+2x+3). ] (The subtitles note the quadratic is not factored further in the video.)

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Speakers / sources featured (as named in the subtitles)

• Dini Handayani (host/presenter)
• “Dini Handayani on the Map channel” (channel/source attribution mentioned in the subtitle)