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When we divide numbers, we often get a quotient and sometimes a remainder. For example, dividing 17 by 5 gives a quotient of 3 and a remainder of 2 because 5 x 3 = 15, and 17 - 15 = 2.
Similarly, in algebra, we can divide polynomials. A polynomial is an expression involving variables and coefficients, such as \( f(x) = x^3 - 4x^2 + 5x - 2 \). When dividing one polynomial by another, we also get a quotient and a remainder.
However, polynomial division can be lengthy and complex. The Remainder Theorem provides a clever shortcut to find the remainder quickly when dividing a polynomial by a simple linear divisor of the form \( (x - a) \). This theorem is especially useful in competitive exams where time is limited.
In this section, we will learn what the Remainder Theorem is, how to apply it, and how it connects to other important concepts like factorization and solving polynomial equations.
Consider a polynomial \( f(x) \) and a divisor of the form \( (x - a) \), where \( a \) is a constant. When we divide \( f(x) \) by \( (x - a) \), the remainder is always a constant (a number without \( x \)).
The Remainder Theorem states:
The remainder when a polynomial \( f(x) \) is divided by \( (x - a) \) is equal to \( f(a) \).
In other words, instead of performing polynomial division, you can simply substitute \( x = a \) into the polynomial and calculate \( f(a) \). The value you get is the remainder.
graph TD A[Start with polynomial f(x)] B[Divide by (x - a)] C[Substitute x = a in f(x)] D[Calculate f(a)] E[Result is remainder R] A --> B --> C --> D --> E
This substitution method saves time and effort, especially in exams.
The Factor Theorem is closely related to the Remainder Theorem and helps us understand when \( (x - a) \) is a factor of the polynomial \( f(x) \).
It states:
If \( f(a) = 0 \), then \( (x - a) \) is a factor of \( f(x) \).
This means that if substituting \( x = a \) into the polynomial gives zero, the polynomial is exactly divisible by \( (x - a) \) with no remainder.
| Condition | Remainder | Conclusion |
|---|---|---|
| \( f(a) eq 0 \) | Non-zero constant | \( (x - a) \) is not a factor |
| \( f(a) = 0 \) | Zero | \( (x - a) \) is a factor |
This theorem is very useful for factorizing polynomials and finding their roots (solutions).
Step 1: Identify \( a = 2 \) from the divisor \( (x - 2) \).
Step 2: Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^3 - 4(2)^2 + 5(2) - 2 \)
\( = 8 - 4 \times 4 + 10 - 2 \)
\( = 8 - 16 + 10 - 2 \)
\( = (8 - 16) + (10 - 2) = -8 + 8 = 0 \)
Answer: The remainder is 0, so \( (x - 2) \) is a factor of \( f(x) \).
Step 1: Identify \( a = 3 \).
Step 2: Calculate \( f(3) \):
\( f(3) = 2(3)^3 - 9(3)^2 + 12(3) - 4 \)
\( = 2 \times 27 - 9 \times 9 + 36 - 4 \)
\( = 54 - 81 + 36 - 4 \)
\( = (54 - 81) + (36 - 4) = -27 + 32 = 5 \)
Step 3: Since \( f(3) eq 0 \), the remainder is 5.
Answer: \( (x - 3) \) is not a factor of \( f(x) \).
Step 1: Identify \( a = 4 \).
Step 2: Calculate \( P(4) \):
\( P(4) = 3(4)^3 - 5(4)^2 + 2(4) + 7 \)
\( = 3 \times 64 - 5 \times 16 + 8 + 7 \)
\( = 192 - 80 + 15 \)
\( = 127 \)
Step 3: The remainder is 127 (thousands of INR).
Interpretation: When the profit polynomial is divided by \( (x - 4) \), the remainder 127 means that at 400 units sold, the profit is INR 127,000.
Step 1: Rewrite divisor \( 2x + 3 \) in the form \( (x - a) \).
\( 2x + 3 = 0 \Rightarrow x = -\frac{3}{2} \)
So, divisor can be written as \( (x - (-\frac{3}{2})) = (x + \frac{3}{2}) \).
Step 2: Substitute \( x = -\frac{3}{2} \) into \( f(x) \):
\( f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^3 + 2\left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right) + 5 \)
\( = -\frac{27}{8} + 2 \times \frac{9}{4} + \frac{3}{2} + 5 \)
\( = -\frac{27}{8} + \frac{18}{4} + \frac{3}{2} + 5 \)
Convert all to eighths:
\( -\frac{27}{8} + \frac{36}{8} + \frac{12}{8} + \frac{40}{8} = \frac{-27 + 36 + 12 + 40}{8} = \frac{61}{8} \)
Answer: The remainder is \( \frac{61}{8} \).
Step 1: Try possible roots using factors of constant term 6: \( \pm1, \pm2, \pm3, \pm6 \).
Step 2: Check \( f(1) \):
\( 1 - 6 + 11 - 6 = 0 \), so \( (x - 1) \) is a factor.
Step 3: Divide \( f(x) \) by \( (x - 1) \) using polynomial division or synthetic division:
Quotient is \( x^2 - 5x + 6 \).
Step 4: Factorize the quadratic:
\( x^2 - 5x + 6 = (x - 2)(x - 3) \).
Answer: \( f(x) = (x - 1)(x - 2)(x - 3) \).
When to use: When the divisor is not in the form \( (x - a) \), such as \( (2x + 3) \).
When to use: When asked to find remainder quickly for polynomial division by linear factors.
When to use: When checking if \( (x - a) \) is a factor of \( f(x) \).
When to use: To speed up calculations in competitive exams.
When to use: When solving polynomial equations or simplifying expressions.
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