Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
Chapter 3
Factorizing a Polynomial
Quadratic Equation
Factorizing by grouping like terms
and Polynomial Factorization
We know that x4 – 6x3 + x – 6 = (x – 6) (x + 1) (x2 – x + 1) now. But how is it found?
First, review some basic facts:
What is Quadratic Equation?
x n = x ⋅ x ⋅ x
⋅ x ⋅ …
⋅x
a (b + c) = ab + ac
n
Quadratic equation (二次方程) is a type of polynomials. It is an equation that the
left-hand-side is a polynomial with degree 2, and usually 1 unknown, and the right-hand-side
is a zero.
We’ll factorize x4 – 6x3 + x + 6 by these facts.
x 4 − 6 x3 + x − 6
= x ⋅ x3 − 6 x3 + x − 6
A general expression of quadratic equation is:
ax2 + bx + c = 0
= x3 ( x − 6 ) + ( x − 6)
= ( x 3 + 1) ( x − 6 )
Where a, b and c are some constants, and x is the variable.
Now we’ve successfully factorized x4 – 6x3 + x – 6 into (x – 6) (x3 + 1) by using the method
“Grouping like terms”.
What is Polynomial Factorization?
Factorizing by identities
Polynomial Factorization (多項式因式分解) is the action that change a polynomial into
products of “smaller” polynomials. For example:
x4 – 6x3 + x – 6
can be factorized into:
(x – 6) (x + 1) (x2 – x + 1)
2
Here, (x + 6), (x + 1) and (x – x + 1) are called “factors” (因子) of x4 – 6x3 + x + 6.
Factorization can be considered as the reverse action of multiplication.
But clearly we haven’t finished. The term (x3 + 1) can be further factored. How do we
factorize it? Grouping terms seems not working! So, instead of grouping terms, we may also
need some help from identities. From chapter 1, we learnt that:
1. (a + b)2 = a2 + 2ab + b2
2. (a – b)2 = a2 – 2ab + b2
3. (a + b)(a – b) = a2 – b2
For example, when we see the expression 4x2 – 1, we knew that it could be factorized into
(2x + 1)(2x – 1), because of identity #3. Here we present two more identities:
5. a3 + b3 = (a + b)(a2 – ab + b2)
6. a3 – b3 = (a – b)(a2 + ab + b2)
Thus, by identity #5, we can reduce (x3 + 1) into (x + 1) (x2 – x + 1), and we have fully
factorized x4 – 6x3 + x – 6.
16
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
Application of Factorization
Solving a Quadratic Equation
One big use of polynomial factorization is to find roots (根) (solutions) of a polynomial. For
example, we need to solve x4 – 6x3 + x – 6 = 0. It is hard to do so. However, we knew that if
some numbers are multiplied and the result is zero, then one of them must be zero. (i.e.,
if a ⋅ b ⋅ c ⋅ d ⋅ e ⋅ f ⋅ … = 0 , then a = 0 or b = 0 or c = 0 or d = 0 or e = 0 or f = 0 or …) This also
applies for polynomials. Thus, if
x4 – 6x3 + x – 6 = 0
Then also
(x – 6) (x + 1) (x2 – x + 1) = 0
That means
x–6=0
or
x+1=0
or
x2 – x + 1 = 0
Solving the first 2 equations gives x = 6 or –1. The last equation gives no real roots (實根) of
x, and we’ll explain why in the later section.
In general, solving a quadratic equation ax2 + bx + c = 0 can be done in several ways.
We will use x2 – 14x + 40 = 0 in our example.
Cross method (十字相乘法
十字相乘法)
十字相乘法
The cross method is to solve a quadratic equation by factorizing.
We may assume that x2 – 14x + 40 = (x – α) (x – β).
Multiplying the right-hand-side, we get x2 – 14x + 40 = x2 – (α + β) x + αβ.
α + β = 14
By comparing coefficients, we get 
. However, method of substitution do not
 αβ = 40
work here, because we don’t know how to solve 14α – α2 = 40 yet. But, we may use a
low-level method to find out α and β: by guessing.
Of course we shouldn’t guess randomly. From the simultaneous equation above, we knew
that αβ = 40, that means, α and β are factors of 40. Therefore, α and β may be 1, 2, 4, 5, 8,
10, 20, 40, -1, -2, -4, -5, -8, -10, -20 or -40.
Another reason for factorizing a polynomial is to reduce computation time. For example,
calculating the polynomial x4 – 6x3 + x – 6 needs 8 multiplications and 3 additions (including
subtraction). In contrast, the factorized form (x – 6) (x + 1) (x2 – x + 1) only needs 3
multiplications and 4 additions. Since time needed for doing multiplications is much more
than additions, computation time is reduced.
When α = …
1
2
4
5
8
10
20
40
-1
-2
-4
-5
-8
-10
-20
-40
Last but not least, factorizing a polynomial is also useful in simplifying an expression. For
x 3 − 6 x 2 + 11x − 6
example,
is a very tedious expression. But if we factorize it, it
3
5 x + 15 x 2 − 170 x + 240
( x − 1)( x − 2 )( x − 3) . The factors (x – 2) and (x – 3) can be cancelled out, and the
becomes
5 ( x + 8 )( x − 2 )( x − 3)
remaining part is just simply
( x − 1) .
5( x + 8)
17
β=…
40
20
10
8
5
4
2
1
-40
-20
-10
-8
-5
-4
-2
-1
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
Next step is to select to right solution, by the condition α + β = 14. (1, 40) is not a solution,
because 1 + 40 = 41 ≠ 14. (2, 20) is not either. (4, 10) is the solution. As soon as we found out
the correct one, we can stop. Therefore, x2 – 14x + 40 = (x – 4) (x – 10), and thus x = 4 or 10.
Sum and Product of Roots of Quadratic Equation
We knew that a quadratic equation ax2 + bx + c could be factorized into (x – α) (x – β). We
also knew that α and β are roots of the quadratic equation. Expand (x – α) (x – β), and by
comparing coefficients, we get:
b

α + β = − a

 αβ = c

a
This can help us find out sum and product of roots without solving the equation.
Completing the square (配方
配方)
配方
Trial and error is easy to follow, but it is extremely inefficient. Imagine if we have a long list
to test, such as x2 + 490x + 5040? You must have been exhausted before you get the answer.
Therefore we need a more systematic way to do so. By our identity #1 and #2, we recognize
that the form x2 ± 2xy + y2 can be collapsed into (x ± y)2. (Here “±” means “+” or “–”.) Can it
be done to our x2 – 14x + 40? Sure. By comparing coefficients, we get x = x (Of course) and
y = 7. (Take “±” as “–”) But where does y2, i.e. 49 come? The answer is easy: add nine and
subtract nine. So:
x 2 − 14 x + 40
After finding the sum and product, we can derive many other forms involving the two roots.
Here gives some important forms composed of roots of a quadratic equation, α and β. For
convenient, s denotes α + β, and p denotes αβ.
= x − 14 x + 40 + 9 − 9
2
= x 2 − 14 x + 49 − 9
α2 + β2
= ( x − 7) − 9
This process is called “Completing the square”. And therefore:
x 2 − 14 x + 40 = 0
2
1.
= α 2 + β 2 + 2αβ − 2αβ
= s2 − 2 p
( x − 7) − 9 = 0
2
( x − 7) = 9
2
α −β
2.
x − 7 = 3 or − 3
x = 10 or 4
= α 2 − 2αβ + β 2 (Assuming α > β)
= s2 − 4 p
Quadratic Formula (二次方程公式
二次方程公式)
二次方程公式
Completing the square is still quite inefficient. Actually, if we complete the square of a
general quadratic equation, ax2 + bx + c = 0, we will get the equation solving a general
quadratic equation, known as “Quadratic formula”:
−b ± b 2 − 4ac
2a
This equation is so important that should be memorized.
x=
18
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
2

 3  1
because x = 0. After completing the square, the equation becomes y =  x −  −   − , and
 2  4

the minimum is…?
For a general quadratic equation, there is many useful information can be found on it.
Let the equation be y = ax2 + bx + c, and the form after completing the square is
y = a(x – h)2 + k. Then:
Discriminant of Quadratic Equation (二次方程之判別式
二次方程之判別式)
二次方程之判別式
−b ± b 2 − 4ac
. We see that the part inside the square root
2a
2
sign, b – 4ac is quite important to the nature of roots. If b2 – 4ac > 0, we got two distinct real
roots. If b2 – 4ac = 0, we got one real root, or two repeated roots (重根). If b2 – 4ac < 0,
there is no real roots.
Recall the quadratic formula x =
Actually, b2 – 4ac is called the “Discriminant” of a quadratic equation: to discriminate the
nature of roots. It is often denoted as D or ∆.
Graph of Quadratic Equation
To begin, see a graph of quadratic equation first:
y
2
y = x^2 + 3x + 2
a controls the shape of the graph.
If a > 0, the graph opens upwards.
If a < 0, the graph opens downwards.
Moreover, the higher |a| is, the “thinner” the graph is.
(0, c) is the y-intercept of the graph.
∆ (Discriminant) controls the relationship between the graph and the x-axis.
If ∆ > 0, the graph intersects the x-axis at two points.
If ∆ = 0, the graph touches, or tangent to (切於) the x-axis at one point.
If ∆ < 0, the graph does not intersect the x-axis.
(0, α), (0, β) are x-intercepts of the graph. (α, β are roots)
k is the extremum (極值) of the graph. Moreover,
If a > 0, it is the minimum.
If a < 0, it is the maximum.
The point (h, k) is the vertex (頂點) of the graph.
Extension: The graph of quadratic equation is called a “parabola” (拋物線)
1
Simultaneous Equations in 2 Unknowns, One Linear, One
Quadratic (聯立二元一次及二次方程
聯立二元一次及二次方程)
聯立二元一次及二次方程
x
-3
-2
-1
0
1
2
A set of simultaneous equations like:
 y = ax 2 + bx + c
…(1)

…( 2 )
 y = mx + k
It can be solved by substituting (2) into (1).
Like quadratic equations, the set can have two solutions, one solution or no solutions.
This is a graph of a quadratic equation, y = x2 + 3x + 2. One can see it crosses the x-axis at
(-2,0) and (-1,0). Also, it crosses the y-axis at (0,2). Anything else special? It achieves its
minimum at ( − 32 , − 41 ) . What is the fact behind these numbers? Let us uncover them. The
x-intercepts are actually the roots, because y = 0 there. The y-intercept is the constant c,
19
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
Factor Theorem (因式定理
因式定理)
因式定理
Remainder Theorem for Quadratic Divisor
Remember the remainder theorem taught in Chapter 1? When a polynomial P(x) is divided by
x – a, its remainder is P(a). But if P(a) = 0, P(x) is divisible by x – a, thus P(x) = (x – a) Q(x)
(Q(x) is the quotient), which implies x – a is a factor of P(x). This gives the factor theorem:
In Chapter 1 Exercise Question 5, you proved that remainder theorem is true because
P ( a ) = ( a − a ) Q ( a ) + R = R . (Oh, I didn’t know you haven’t done that)
By similar method, we can also deduce a “remainder theorem” whose divisor is quadratic.
If P(x) is a polynomial and P(a) = 0, x – a is a factor of P(x).
Assume the divisor is ax2 + bx + c, and dividend is P(x). Thus,
P ( x ) = ( ax 2 + bx + c ) Q ( x ) + R ( x )
For example, to factorize P(x) = x3 + 6x2 + 11x + 6, we may employ the factor theorem and
test for every possible a so that P(a) = 0. From our experience in doing cross method for
quadratic equation, we could also list out all factors of 6 and test for them one by one.
Thus, P(x) = (x + 1)(x + 2)(x + 3). Except testing out all possibilities, we may also test the
possibilities until one zero is found (e.g. a = -1 above), and do synthetic division to get an
equation with fewer degree (e.g. x2 + 5x + 6 here, as x3 + 6x2 + 11x + 6 = (x + 1)( x2 + 5x + 6))
Once we got a quadratic equation left, instantly apply quadratic formula.
Here, define R(x) = rx + s.
To make the Q(x) part disappear, one must have Q(x) = 0 or ax2 + bx + c = 0. We chose the
latter one, as it is easier to implement. ax2 + bx + c = 0 when x = α or x = β, where α, β are
roots. So we have got:
 P (α ) = R (α )

 P ( β ) = R ( β )
Or
 P (α ) = rα + s

 P ( β ) = r β + s
An SLE2!
P (α ) − P ( β )
. Substitute this into any of the equation above to get
Solving for r, we got r =
α −β
the value of s.
By this, we got the remainder theorem for quadratic divisor.
A
more
complicated
example
is
that
factorizing
the
polynomial
P5 ( x ) = x 5 − 2 x 4 − 39 x 3 + 92 x 2 + 188 x − 240 . Our list is quite long, 40 candidates. So, by
Revision:
a
1
2
3
6
-1
-2
-3
-6
P(x)
24
60
120
504
0
0
0
-60
trial and error, we got (x – 1) as one of the factors. Dividing, our new polynomial is
P4 ( x ) = x 4 − x 3 − 40 x 2 + 52 x + 240 . Still that 40 values to be tested, and we have (x + 2) this
In this chapter, we’ve learnt:
1. What is, how to, and why factorizing a polynomial
2. Solving quadratic equation
3. Sum and product of roots of quadratic equation
4. Discriminant of quadratic equation
5. Simultaneous equations in 2 unknowns, one linear, one quadratic
6. Factor theorem
7. Remainder theorem for quadratic divisor
time. P3 ( x ) = x 3 − 3 x 2 − 34 x + 120 , and the factor is (x – 4). What we left is only
P2 ( x ) = x 2 + x − 30 , so apply quadratic formula to get (x – 5) and (x + 6). Therefore,
x 5 − 2 x 4 − 39 x 3 + 92 x 2 + 188 x − 240 = ( x − 1)( x + 2 )( x − 4 )( x − 5 )( x + 6 ) . Tedious, is it?
When using factor theorem, you should first test the values that has a small absolute value.
20
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
2
 y = 3 − x − 2 x
10. Solve 
 0 = x − 2 y + 5
Exercise
11. (ISMC 2000 Final) If a + 1 + b 2 + 1 + 2b = 0 , find a2000 + b2001
In the followings, if not specified, x is the variable, k is a constant, and n is an integer.
[Hint: k ≥ 0 ]
12. (PCMSIMC 2003, Modified) Let f(x) = ax + b2. If f(f(x)) = 4x + 9,
a) Find all possible values of a and b.
b) Find the smallest possible value of f(1).
13. Assume A (k, k2), B (1, k), C (k2, 1) are three points on the coordinates plane, where k ≠ 1.
a) Express the area of ∆ABC in terms of k.
b) Find the area of ∆ABC when k = 3.
c) Find the integral solution (整數解) of k when the area of ∆ABC is 3.5.
14. (USAMO 1995, Modified) If f(x) = ax4 – bx2 + x + 5 and f(-3) = 2, show that f(x) is not
divisible by x2 – 9.
15. Prove the properties of sum and product of roots by quadratic formula.
x2
16. (CaMO 1992) Solve x 2 +
=3
2
( x + 1)
1. Solve the following equations. (If there is no real roots, state that)
a) x2 + 6x + 5 = 0
b) 2x2 + 30x + 112 = 0
c) x2 + 4x + 1 = 0
d) x2 + x + 1 = 0
e) x4 - 9x2 + 8 = 0
f) x10 – 21x9 + 110x8 = 0
x
2+ x
=
g)
1 + 2x
x
x − 3 x −1 −1 = 2
h)
Solve x3 – 14x2 + 59x – 70 = 0
If α and β are roots of x2 + 2x + 3, find α3 + β3.
Factorize x6 + x5 – 41x4 + 105x3 – 106x2 + 104x – 64.
If x2 + kx + k = 1 has repeated roots for x, find k.
Refer to the followings.
a) Find 220.
x10 − 3 x − 4
b) Find the remainder of 2
.
x − 3x − 4
7. (HKCEE 1988) If 9x2 – (k+1)x + 1 = 0 has equal roots,
a) Solve k.
b) Solve the equation if k < 0.
8. (HKCEE 1994) Solve (x – 3)2 – |x – 3| – 12 = 0.
9. (HKCEE 1996) Let α, β be real roots of x2 – λx + 1 = 0, where λ > 2.
Let sn = αn + βn, where n is a positive integer.
a) Express s2 and s3 in terms of λ.
b) Find a5 – λα4 + α3.
c) Find s5 – λs4 + s3.
d) It is known that s3, s4, s5 ≠ 0 and s3 : s4 : s5 = 10 : 7λ : 25. Find λ.
e) Find s3.
2.
3.
4.
5.
6.
5
 5 + 1  5 −1
f) Evaluate 
 + 

 2   2 
17. Factorize x5 + 4x4 + 10x3 + 16x2 + 17x + 12
[Hint: This is product of a quadratic and cubic equation… How did cross method work?]
18. (ISMC 2000 Final) Factorize z6 + z4 + z3 + z2 + 1.
[Hint: This is product of a quadratic and quartic equation]
5
21
Advanced Mathematics Training Class Notes
Chapter 3: Quadratic Equation and Polynomial Factorization
12a) a = 2, b = − 2; a = −2, b = 2
b) 1
Suggested Solutions for the Exercise
13a) 12 k 4 − k 3 − k + 1
1a) –5, -1
b) –8, -7
c) −2 ± 3
d) No real roots
e) −1, 1, − 2 2, 2 2
f) 0, 10, 11
( k − 1)
1± 5
2
17) (x2 + 2x + 3) (x3 + 2x2 + 3x + 4)
18) (z2 – z + 1) (z4 + z3 + z2 + z + 1)
16)
5 ± 17
2
h) 17
2) 2, 5, 7
3) 10
4) (x – 1) (x – 2) (x – 4) (x + 8) (x2 + 1)
5) 2
6a) 1048576
b) 209712x + 209712
7a) 5, -7
b) –1/3
8) –1, 7
9a) s2 = λ2 – 2; s3 = λ3 – 3λ.
b) 0
c) 0
d) 5
e) 2 5
f) 5 5
5 ± 33
;
4
1
2
b) 26
c) 2
g) −
10) x = −
or
y=
15 ∓ 33
8
11) 0
22
2
(k
2
+ k + 1)