Math Problem Statement

Suku banyak berderajat tiga P(x) = x^3 + 2x^2 + ax + b dibagi dengan x^2 - 3x + 2 mempunyai sisa 3x + 2, nilai a + b = ...

Solution

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Suku banyak berderajat tiga
P(x)=x3+2x2+ax+bP(x) = x^3 + 2x^2 + ax + b
dibagi dengan x23x+2x^2 - 3x + 2 mempunyai sisa 3x+23x + 2, nilai a+b=a + b = \ldots


The problem asks for the value of a+ba + b when the polynomial P(x)=x3+2x2+ax+bP(x) = x^3 + 2x^2 + ax + b is divided by x23x+2x^2 - 3x + 2 and gives a remainder of 3x+23x + 2.

Let's solve this step by step.

Step 1: Polynomial Division Setup

When a polynomial P(x)P(x) is divided by D(x)=x23x+2D(x) = x^2 - 3x + 2, the division yields a quotient Q(x)Q(x) and a remainder R(x)R(x). So, we can express it as:

P(x)=(x23x+2)Q(x)+(3x+2)P(x) = (x^2 - 3x + 2)Q(x) + (3x + 2)

Step 2: Assume the Form of the Quotient Q(x)Q(x)

The degree of P(x)P(x) is 3, and D(x)D(x) has degree 2. Therefore, Q(x)Q(x) should be a linear polynomial of the form Q(x)=cx+dQ(x) = cx + d.

Step 3: Expand the Expression

Now, let's substitute and expand:

P(x)=(x23x+2)(cx+d)+(3x+2)P(x) = (x^2 - 3x + 2)(cx + d) + (3x + 2)

Expand the product:

(x23x+2)(cx+d)=x2(cx)+x2(d)3x(cx)3x(d)+2(cx)+2(d)(x^2 - 3x + 2)(cx + d) = x^2(cx) + x^2(d) - 3x(cx) - 3x(d) + 2(cx) + 2(d) =cx3+dx23cx23dx+2cx+2d= cx^3 + dx^2 - 3cx^2 - 3dx + 2cx + 2d =cx3+(d3c)x2+(3d+2c)x+2d= cx^3 + (d - 3c)x^2 + (-3d + 2c)x + 2d

Step 4: Combine with the Remainder

Now, we combine this with the remainder:

P(x)=cx3+(d3c)x2+(3d+2c)x+2d+3x+2P(x) = cx^3 + (d - 3c)x^2 + (-3d + 2c)x + 2d + 3x + 2

Step 5: Compare Coefficients

We are given that P(x)=x3+2x2+ax+bP(x) = x^3 + 2x^2 + ax + b. Therefore, equate the corresponding coefficients:

  • For x3x^3: c=1c = 1
  • For x2x^2: d3c=2d - 3c = 2
    Substituting c=1c = 1:
    d3(1)=2d3=2d=5d - 3(1) = 2 \Rightarrow d - 3 = 2 \Rightarrow d = 5
  • For xx: 3d+2c+3=a-3d + 2c + 3 = a
    Substituting d=5d = 5 and c=1c = 1:
    3(5)+2(1)+3=a15+2+3=aa=10-3(5) + 2(1) + 3 = a \Rightarrow -15 + 2 + 3 = a \Rightarrow a = -10
  • For the constant: 2d+2=b2d + 2 = b
    Substituting d=5d = 5:
    2(5)+2=b10+2=bb=122(5) + 2 = b \Rightarrow 10 + 2 = b \Rightarrow b = 12

Step 6: Find a+ba + b

Now, a=10a = -10 and b=12b = 12, so:

a+b=10+12=2a + b = -10 + 12 = 2

Final Answer:

The value of a+ba + b is 2\boxed{2}.

Would you like a more detailed explanation, or do you have any questions?


Here are 5 related questions:

  1. What is the degree of a polynomial and how does it affect division?
  2. How do you perform polynomial long division?
  3. What is the significance of the remainder when dividing polynomials?
  4. Can we always determine the quotient's degree from the degrees of the dividend and divisor?
  5. How would the solution change if the divisor had a higher degree?

Tip: When dividing polynomials, comparing coefficients is a powerful method to solve for unknowns efficiently!

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Math Problem Analysis

Mathematical Concepts

Polynomial Division
Algebra

Formulas

P(x) = (x^2 - 3x + 2) * Q(x) + (3x + 2)

Theorems

Polynomial Division Theorem

Suitable Grade Level

Grades 10-12