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Functions - Polynomial Functions (Factor and Remainder Theorems)

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Polynomial Definition: A polynomial P(x)P(x) of degree nn is defined as P(x)=anxn+anβˆ’1xnβˆ’1+β‹―+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0, where anβ‰ 0a_n \neq 0. Visually, the graph of a polynomial is a continuous, smooth curve without breaks or sharp corners. The degree nn determines the maximum number of xx-intercepts and the general 'end behavior' of the graph as xx approaches infinity.

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The Remainder Theorem: This theorem states that when a polynomial P(x)P(x) is divided by a linear factor (xβˆ’c)(x - c), the remainder is simply the value of the polynomial evaluated at cc, or R=P(c)R = P(c). This allows for quick calculation of remainders without performing full algebraic long division.

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The Factor Theorem: A specific case of the Remainder Theorem, it states that (xβˆ’c)(x - c) is a factor of P(x)P(x) if and only if P(c)=0P(c) = 0. Visually, this means that if (xβˆ’c)(x - c) is a factor, the graph of y=P(x)y = P(x) will have an xx-intercept at the point (c,0)(c, 0).

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Division Algorithm: Any polynomial P(x)P(x) can be expressed in terms of a divisor D(x)D(x), a quotient Q(x)Q(x), and a remainder R(x)R(x) such that P(x)=D(x)Q(x)+R(x)P(x) = D(x)Q(x) + R(x). In the context of the Factor Theorem, if the remainder is zero, the divisor and quotient are both factors of the original polynomial.

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Linear Divisors of the form (axβˆ’b)(ax - b): When a polynomial is divided by (axβˆ’b)(ax - b), the Remainder Theorem can be extended to state that the remainder is P(ba)P(\frac{b}{a}). If P(ba)=0P(\frac{b}{a}) = 0, then (axβˆ’b)(ax - b) is a factor of the polynomial.

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Roots and Multiplicity: If (xβˆ’c)k(x - c)^k is a factor of P(x)P(x), then cc is a root of multiplicity kk. Visually, if k=1k=1, the graph crosses the xx-axis at cc. If kk is even (e.g., k=2k=2), the graph is tangent to the xx-axis at cc (it touches and turns back, forming a 'U' or inverted 'U' shape). If kk is odd and greater than 1, the graph flattens as it crosses the axis.

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Finding Unknown Coefficients: By using given information about factors or remainders, we can create a system of linear equations to solve for unknown coefficients (like kk or mm) within a polynomial expression. For example, knowing that (xβˆ’1)(x-1) and (x+2)(x+2) are factors provides two equations: P(1)=0P(1)=0 and P(βˆ’2)=0P(-2)=0.

πŸ“Formulae

P(x)=anxn+anβˆ’1xnβˆ’1+β‹―+a1x+a0P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0

P(x)=(xβˆ’c)Q(x)+RP(x) = (x - c)Q(x) + R

R=P(c)Β whenΒ dividingΒ byΒ (xβˆ’c)R = P(c) \text{ when dividing by } (x - c)

R=P(ba)Β whenΒ dividingΒ byΒ (axβˆ’b)R = P\left(\frac{b}{a}\right) \text{ when dividing by } (ax - b)

IfΒ P(c)=0β€…β€ŠβŸΉβ€…β€Š(xβˆ’c)Β isΒ aΒ factorΒ ofΒ P(x)\text{If } P(c) = 0 \implies (x - c) \text{ is a factor of } P(x)

πŸ’‘Examples

Problem 1:

Find the remainder when P(x)=2x3βˆ’5x2+xβˆ’7P(x) = 2x^3 - 5x^2 + x - 7 is divided by (xβˆ’3)(x - 3).

Solution:

Step 1: Identify the value of cc from the divisor (xβˆ’3)(x - 3). Here, c=3c = 3. Step 2: Apply the Remainder Theorem, which states R=P(3)R = P(3). Step 3: Substitute x=3x = 3 into the polynomial: P(3)=2(3)3βˆ’5(3)2+(3)βˆ’7P(3) = 2(3)^3 - 5(3)^2 + (3) - 7 Step 4: Calculate the powers: P(3)=2(27)βˆ’5(9)+3βˆ’7P(3) = 2(27) - 5(9) + 3 - 7 Step 5: Simplify the expression: P(3)=54βˆ’45+3βˆ’7=5P(3) = 54 - 45 + 3 - 7 = 5 The remainder is 55.

Explanation:

We use the Remainder Theorem to avoid long division. Since we are dividing by (xβˆ’3)(x - 3), evaluating the polynomial at x=3x = 3 gives the remainder directly.

Problem 2:

The polynomial P(x)=x3+kx2βˆ’4x+12P(x) = x^3 + kx^2 - 4x + 12 has (x+2)(x + 2) as a factor. Determine the value of kk and find the other factors.

Solution:

Step 1: Since (x+2)(x + 2) is a factor, by the Factor Theorem, P(βˆ’2)=0P(-2) = 0. Step 2: Substitute x=βˆ’2x = -2 into P(x)P(x) and set it to zero: (βˆ’2)3+k(βˆ’2)2βˆ’4(βˆ’2)+12=0(-2)^3 + k(-2)^2 - 4(-2) + 12 = 0 Step 3: Solve for kk: βˆ’8+4k+8+12=0-8 + 4k + 8 + 12 = 0 4k+12=0β€…β€ŠβŸΉβ€…β€Š4k=βˆ’12β€…β€ŠβŸΉβ€…β€Šk=βˆ’34k + 12 = 0 \implies 4k = -12 \implies k = -3 Step 4: Rewrite the polynomial with k=βˆ’3k = -3: P(x)=x3βˆ’3x2βˆ’4x+12P(x) = x^3 - 3x^2 - 4x + 12 Step 5: Perform division or factor by grouping to find the remaining factors: x2(xβˆ’3)βˆ’4(xβˆ’3)=(x2βˆ’4)(xβˆ’3)x^2(x - 3) - 4(x - 3) = (x^2 - 4)(x - 3) Step 6: Factor the difference of squares: (xβˆ’2)(x+2)(xβˆ’3)(x - 2)(x + 2)(x - 3) The value of kk is βˆ’3-3 and the other factors are (xβˆ’2)(x - 2) and (xβˆ’3)(x - 3).

Explanation:

We first apply the Factor Theorem to solve for the unknown coefficient kk. Once the polynomial is complete, we use algebraic factoring techniques (grouping) to decompose the cubic into its linear factors.