Chapter 2: Polynomials
Polynomial division, synthetic division, remainder theorem, factor theorem, factorization and polynomial equations — reconstructed from the supplied 39-page handwritten notes.
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Introduction to Polynomials
The word polynomial comes from two words: poly, meaning “many,” and nomial, meaning “terms.” Thus, a polynomial is an algebraic expression consisting of one or more terms with non-negative integral powers of the variable.
Division Algorithm of Polynomials
If a polynomial f(m) is divided by a polynomial g(m), then:
- f(m) = Dividend
- g(m) = Divisor
- Q(m) = Quotient
- R(m) = Remainder
Example
Divide 2m³ − 5m² + 5m − 3 by 2m − 3.
The handwritten source works this first by general polynomial division and then by synthetic division on page 2.
Synthetic Division
For a linear divisor, first write the value that makes the divisor zero. For example, for:
Using synthetic division for the coefficients of 2m³ − 5m² + 5m − 3, the source obtains zero remainder.
Exercise 2.1
1. Divide the Polynomial by Synthetic Division
(a) f(m)=2m²−m−3, g(m)=2m−3
2m−3=0 ⇒ m=3/2
Using synthetic division, the remainder is 0.
The source writes the quotient in equivalent scaled form and concludes the division is exact.
(b) f(m)=12m³−20m²−9m+15, g(m)=3m−5
3m−5=0 ⇒ m=5/3
Synthetic division gives remainder 0.
The source quotient is equivalent to:
4m²−5m−3
2. Find the Quotient Q(m) and Remainder R(m) by General Division
(a) m³−9m²+17m−4 divided by (m−2)
The source performs long division and obtains:
Q(m)=m²−7m+3
R(m)=2
(b) 4m³+4m²−2m+16 divided by (2m−3)
After long division, the source records:
Q(m)=2m²+5m+5
R(m)=2
(c) 3m⁴−8m²−20 divided by (m²−2)
The source obtains:
Q(m)=3m²−2
R(m)=−24
3. Find the Polynomial f(m) When Q(m), g(m) and R(m) are Given
Use:
(a)
Q(m)=m+3, g(m)=m−1 and R(m)=5
f(m)=(m−1)(m+3)+5
=m²+2m−3+5
∴ f(m)=m²+2m+2
(b)
Q(m)=3m+5, g(m)=3m−5 and R(m)=7
f(m)=(3m−5)(3m+5)+7
=9m²−25+7
∴ f(m)=9m²−18
(c)
The source substitutes the given Q(m), g(m) and R(m), then simplifies to:
f(m)=−m³+4m²−4m+8
(d)
The handwritten source obtains a cubic polynomial by multiplying the given divisor and quotient and adding the remainder.
5. Find the Quotient and Remainder by Synthetic Division
(a) m³+6m²−m−30 divided by (m−2)
m−2=0 ⇒ m=2
Synthetic division gives:
Q(m)=m²+8m+15
R(m)=0
(b) Polynomial divided by (m+2)
m+2=0 ⇒ m=−2
The source gives:
Q(m)=m³+2m²−9m+18
R(m)=−36
(c) 2m³+3m²−4m−5 divided by (2m−1)
2m−1=0 ⇒ m=1/2
The source obtains zero remainder and writes the quotient in scaled form.
(d) 2m³+5m²+5m−5 divided by (2m+3)
2m+3=0 ⇒ m=−3/2
The source records:
R(m)=−8
and the corresponding quotient from the synthetic row.
6. Using Synthetic Division, Find the Quotient and Remainder
(a) 4m³−2m²−3m+2 divided by (m−2)
m=2
Q(m)=4m²+6m+9
R(m)=20
(b) 2m⁴−7m³+m+12 divided by (m−3)
m=3
The source records:
Q(m)=2m³−m²−3m−8
R(m)=−36
(c) m³−2m+3 divided by (m−1)
m=1
Q(m)=m²+m−1
R(m)=2
(d) m⁴−3m²+6 divided by (m+2)
m=−2
Q(m)=m³−2m²+m−2
R(m)=10
(e) m⁵+m³−5 divided by (m+1)
m=−1
The source gives:
Q(m)=m⁴−m³+2m²−2m+2
R(m)=−7
(g) 4m³+2m²−4m+3 divided by (2m+3)
m=−3/2
The synthetic division gives R(m)=0.
(h) 2m³−9m²+5m−5 divided by (2m−3)
m=3/2
The source obtains:
R(m)=−11
with quotient from the synthetic row.
Remainder Theorem and Factor Theorem
Remainder Theorem
If a polynomial f(m) is divided by (m−a), then the remainder is equal to f(a).
Proof
By the division algorithm:
Putting m=a:
Factor Theorem
If f(a)=0, then (m−a) is a factor of f(m).
Converse of Factor Theorem
If (m−a) is a factor of f(m), then:
Exercise 2.2
1. Find the Remainder Using the Remainder Theorem
(a) 2m³−7m²+5m+4 divided by (m−3)
m−3=0 ⇒ m=3
R=f(3)
=2(3³)−7(3²)+5(3)+4
=54−63+15+4
∴ R=10
(b)
The source evaluates f(−2/3) for the given divisor and concludes:
(c) 4m³−3m²+2m−9 divided by (m+1)
m=−1
R=f(−1)
=−4−3−2−9
∴ R=−18
(d)
For a divisor equivalent to m−3/2, the source evaluates f(3/2) and records:
(e) 3m³+5m²−m−9 divided by (3m+1)
3m+1=0 ⇒ m=−1/3
R=f(−1/3)
The handwritten result is:
R=−29/3
2. Using Factor Theorem, Show that the Given Linear Expression is a Factor
(a) Show (m−1) is a factor
Put m=1.
The source evaluates f(1)=0.
∴ (m−1) is a factor.
(b) Show (m−3) is a factor
Put m=3.
The source evaluates f(3)=0.
∴ (m−3) is a factor.
(c) Show (m+1) is a factor
Put m=−1.
The source evaluates f(−1)=0.
∴ (m+1) is a factor.
(d) Show (m+2) is a factor
Put m=−2.
The source obtains f(−2)=0.
∴ (m+2) is a factor.
(e) Show (2m−3) is a factor
2m−3=0 ⇒ m=3/2
The source evaluates f(3/2)=0.
∴ (2m−3) is a factor.
(f) Show (2m+1) is a factor
2m+1=0 ⇒ m=−1/2
The source evaluates f(−1/2)=0.
∴ (2m+1) is a factor.
5. Find the Value of k
(a) If (2m−5) is a factor of 6m³−(k+6)m²+2km−25
2m−5=0 ⇒ m=5/2
Using the factor theorem, f(5/2)=0.
The source simplifies the resulting equation and obtains:
∴ k=−25/6
(b) If (m−3) is a factor
Put m=3.
The handwritten solution gives:
∴ k=−6
(c) If (m−1) is a factor
Put m=1.
The source obtains:
∴ k=2
(d) If (2m+1) is a factor of 2m³+km²+m+2
2m+1=0 ⇒ m=−1/2
Using f(−1/2)=0, the source obtains:
∴ k=−5
Long Type Questions – Find a and b
The source forms two equations by using two given linear factors. For each factor, the corresponding root is substituted into the polynomial and set equal to zero.
Example pattern
Pages 29–32 solve simultaneous equations in a and b. In one worked problem, the source concludes:
Factorize the Following Polynomials
(a) m³+3m²−4m−12
Possible integral roots are tested using the factor theorem.
f(−2)=0, so (m+2) is a factor.
Synthetic division gives the quotient:
m²+m−6=(m+3)(m−2)
Therefore:
∴ m³+3m²−4m−12=(m+2)(m+3)(m−2)
(b) 2m³−m²−10m−15
The source tests m=3 and obtains f(3)=0.
So (m−3) is a factor.
The remaining quadratic is:
2m²+5m+5
(c) 2m⁴−5m²+3
f(1)=0, so (m−1) is a factor.
The source continues synthetic division and then tests another root.
The factorization proceeds through the quotient shown on pages 35–36.
(d)
Pages 36–38 factor a fourth-degree polynomial by successively testing integer roots and using synthetic division. The handwritten source identifies two linear factors from roots 2 and 3 before completing the factorization.
3. Solve the Polynomial Equation
m³−7m²+7m+15=0
The source tests m=3:
3³−7(3²)+7(3)+15=0
So (m−3) is a factor.
Synthetic division gives:
m²−4m−5
Factorizing the quadratic:
m²−4m−5=(m+1)(m−5)
Thus:
(m−3)(m+1)(m−5)=0
∴ m=−1, 3, 5
Discussion
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