Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Polynomials
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Polynomials
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Subject: Further Mathematics
Class: Senior Secondary 2
Term: 1st Term
Week: 1
Theme: Pure Mathematics
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Define polynomials Divide a polynomial by a polynomial of lesser degree Use remainder the or em Find factor of polynomials Use roots of cubic equation to solve problem on cubic equations
Dividing a polynomial $P(x)$ by another polynomial $D(x)$ (where $D(x)$ is of lesser degree than $P(x)$ and $D(x) \neq 0$) results in a quotient $Q(x)$ and a remainder $R(x)$.
This can be expressed as: $\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$ or $P(x) = D(x) \cdot Q(x) + R(x)$ The degree of the remainder $R(x)$ must be less than the degree of the divisor $D(x)$. If $R(x) = 0$, then $D(x)$ is a factor of $P(x)$.
Method: Long Division This method is similar to numerical long division.
Worked Example 1: Divide $P(x) = 3x^3 + 2x^2 - 10x + 4$ by $D(x) = x - 2$.
Step-by-step Solution: Set up the division: ``` _________ x - 2 | 3x3 + 2x2 - 10x + 4 ``` Divide the leading term of the dividend ($3x^3$) by the leading term of the divisor ($x$): $3x^3 / x = 3x^2$. Write this in the quotient. ``` 3x2 _______ x - 2 | 3x3 + 2x2 - 10x + 4 ``` Multiply the quotient term ($3x^2$) by the entire divisor ($x - 2$): $3x^2(x - 2) = 3x^3 - 6x^2$. Write this below the dividend. ``` 3x2 _______ x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ ``` Subtract the result from the dividend: $(3x^3 + 2x^2) - (3x^3 - 6x^2) = 8x^2$. Bring down the next term ($-10x$). ``` 3x2 _______ x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ 8x2 - 10x ``` Repeat the process with the new polynomial ($8x^2 - 10x$): Divide the leading term ($8x^2$) by $x$: $8x^2 / x = 8x$. Add this to the quotient. ``` 3x2 + 8x _____ x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ 8x2 - 10x -(8x2 - 16x) __________ ``` Subtract: $(8x^2 - 10x) - (8x^2 - 16x) = 6x$. Bring down the next term ($+4$). ``` 3x2 + 8x _____ x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ 8x2 - 10x -(8x2 - 16x) __________ 6x + 4 ``` Repeat again with $6x + 4$: Divide $6x$ by $x$: $6x / x = 6$. Add this to the quotient. ``` 3x2 + 8x + 6 x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ 8x2 - 10x -(8x2 - 16x) __________ 6x + 4 -(6x - 12) __________ ``` Subtract: $(6x + 4) - (6x - 12) = 16$. This is the remainder. The degree of 16 (0) is less than the degree of $x-2$ (1). ``` 3x2 + 8x + 6 x - 2 | 3x3 + 2x2 - 10x + 4 -(3x3 - 6x2) __________ 8x2 - 10x -(8x2 - 16x) __________ 6x + 4 -(6x - 12) __________ 16 ``` Therefore, $Q(x) = 3x^2 + 8x + 6$ and $R(x) = 16$. So, $3x^3 + 2x^2 - 10x + 4 = (x - 2)(3x^2 + 8x + 6) + 16$.
Statement: If a polynomial $P(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is $P(a)$.
Justification/Proof: From the division algorithm, we know that $P(x) = (x - a) Q(x) + R(x)$, where $Q(x)$ is the quotient and $R(x)$ is the remainder. Since the divisor $(x-a)$ has degree 1, the remainder $R(x)$ must have a degree less than 1, meaning $R(x)$ is a constant, say $R$. So, $P(x) = (x - a) Q(x) + R$. Now, substitute $x = a$ into the equation: $P(a) = (a - a) Q(a) + R$ $P(a) = (0) Q(a) + R$ $P(a) = R$ Thus, the remainder when $P(x)$ is divided by $(x - a)$ is $P(a)$.
Worked Example 2: Find the remainder when $f(x) = x^3 - 4x^2 + 5x + 1$ is divided by $(x - 1)$.
Step-by-step Solution: Identify the divisor as $(x - a)$, so $a = 1$. Substitute $a=1$ into $f(x)$ to find $f(1)$. $f(1) = (1)^3 - 4(1)^2 + 5(1) + 1$ $f(1) = 1 - 4 + 5 + 1$ $f(1) = 3$ Therefore, the remainder is
3. Worked Example 3: Find the remainder when $f(x) = 2x^4 - 3x^2 + x - 5$ is divided by $(2x + 1)$.
Step-by-step Solution: The divisor is $2x + 1$. To use the Remainder Theorem, we need it in the form $(x - a)$. Set $2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$. So, $a = -\frac{1}{2}$. Substitute $a = -\frac{1}{2}$ into $f(x)$. $f(-\frac{1}{2}) = 2(-\frac{1}{2})^4 - 3(-\frac{1}{2})^2 + (-\frac{1}{2}) - 5$ $f(-\frac{1}{2}) = 2(\frac{1}{16}) - 3(\frac{1}{4}) - \frac{1}{2} - 5$ $f(-\frac{1}{2}) = \frac{1}{8} - \frac{3}{4} - \frac{1}{2} - 5$ Find a common denominator (8): $f(-\frac{1}{2}) = \frac{1}{8} - \frac{6}{8} - \frac{4}{8} - \frac{40}{8}$ $f(-\frac{1}{2}) = \frac{1 - 6 - 4 - 40}{8}$ $f(-\frac{1}{2}) = \frac{-49}{8}$ Therefore, the remainder is $-\frac{49}{8}$.
Statement: A polynomial $P(x)$ has a factor $(x - a)$ if and only if $P(a) = 0$.
Justification: This is a direct consequence of the Remainder Theorem. If $P(a) = 0$, then the remainder when $P(x)$ is divided by $(x - a)$ is
0. This means $(x - a)$ divides $P(x)$ exactly, making $(x - a)$ a factor of $P(x)$. Conversely, if $(x - a)$ is a factor of $P(x)$, then $P(x) = (x - a)Q(x)$ for some polynomial $Q(x)$. Substituting $x=a$ gives $P(a) = (a-a)Q(a) = 0 \cdot Q(a) = 0$. To factorize a polynomial using the Factor Theorem: Find a value 'a' such that $P(a) = 0$. This implies $(x-a)$ is a factor. Use long division or synthetic division to divide $P(x)$ by $(x-a)$ to find the other factor (the quotient). Factorize the quotient further if it's a quadratic or higher-degree polynomial.
Worked Example 4: Factorize $P(x) = x^3 - 2x^2 - 5x + 6$ completely.
Step-by-step Solution: Find a root: Test integer factors of the constant term (6): $\pm 1, \pm 2, \pm 3, \pm 6$. Test $x=1$: $P(1) = (1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$. Since $P(1) = 0$, $(x - 1)$ is a factor. Divide $P(x)$ by $(x - 1)$ using long division: ``` x2 - x - 6 x - 1 | x3 - 2x2 - 5x + 6 -(x3 - x2) _________ -x2 - 5x -(-x2 + x) _________ -6x + 6 -(-6x + 6) _________ 0 ``` The quotient is $Q(x) = x^2 - x - 6$.
Factorize the quadratic quotient: $x^2 - x - 6 = (x - 3)(x + 2)$ Complete factorization: $P(x) = (x - 1)(x - 3)(x + 2)$ For a cubic equation $ax^3 + bx^2 + cx + d = 0$, if $\alpha$, $\beta$, and $\gamma$ are its roots, then: Sum of roots: $\alpha + \beta + \gamma = -\frac{b}{a}$ Sum of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ Product of roots: $\alpha\beta\gamma = -\frac{d}{a}$ These relationships are useful for solving problems related to cubic equations, especially when one or more roots are known, or when properties of the roots are given.
Worked Example 5: If $x = 2$ is a root of the equation $2x^3 - 7x^2 + kx - 2 = 0$, find the value of $k$ and the other roots.
Step-by-step Solution: Find $k$ using the fact that $x=2$ is a root: If $x=2$ is a root, then substituting $x=2$ into the equation must satisfy it. $2(2)^3 - 7(2)^2 + k(2) - 2 = 0$ $2(8) - 7(4) + 2k - 2 = 0$ $16 - 28 + 2k - 2 = 0$ $-14 + 2k = 0$ $2k = 14$ $k = 7$ Rewrite the equation with the value of $k$: $2x^3 - 7x^2 + 7x - 2 = 0$ Find the other roots: Since $x=2$ is a root, $(x-2)$ is a factor. Divide the polynomial by $(x-2)$. ``` 2x2 - 3x + 1 x - 2 | 2x3 - 7x2 + 7x - 2 -(2x3 - 4x2) __________ -3x2 + 7x -(-3x2 + 6x) __________ x - 2 -(x - 2) ________ 0 ``` The quotient is $2x^2 - 3x + 1$. Solve the quadratic quotient for the remaining roots: $2x^2 - 3x + 1 = 0$ This can be factored: $(2x - 1)(x - 1) = 0$ So, $2x - 1 = 0 \implies x = \frac{1}{2}$ And $x - 1 = 0 \implies x = 1$ Conclusion: The value of $k$ is 7, and the other roots are $x = \frac{1}{2}$ and $x = 1$. The roots of the equation are $2, 1, \frac{1}{2}$.
Engineering and Architecture: Polynomials are used to design shapes and curves in engineering structures, bridges, and architectural designs. For instance, a civil engineer planning the slope of a road or the curve of a bridge arch in a Nigerian city might use cubic or quartic polynomial functions to ensure stability, optimize material use, and achieve desired aesthetics. The roots of polynomial equations in this context could represent critical points or boundaries of the design.
Economics and Business: Businesses and government agencies in Nigeria use polynomial functions to model economic trends, predict market behavior, or forecast commodity prices (e.g., crude oil, agricultural produce like cassava or cocoa). A polynomial could represent the relationship between price, demand, and supply over time. The roots of these polynomials might indicate equilibrium points or periods of zero growth/profit, which are crucial for decision-making by Nigerian entrepreneurs or economic planners.
Science and Environmental Modeling: In fields like epidemiology or environmental science, polynomials can model the growth of populations (human, animal, or even disease outbreaks) or the spread of pollutants. For example, a polynomial might describe the population of a specific fish species in Lake Chad over several years, or the spread of a localized waste spill in the Niger Delta. Solving these polynomial equations could help identify peak populations, extinction points, or critical thresholds for environmental intervention.