Lecture 6 : Some Methods of Factorization

Basic Methods of Factorization

(I) Extract the common factors from terms: like
$xm + ym + zm = m(x + y + z).$

(II) Apply multiplication formulae: like those mentioned in Lecture 5. How- ever, contrary to Lecture 5, at present each formula is applied for con- verting an expression in non-product form to a new expression in product form.

(III) Cross-Multiplication:

$\begin{aligned}
&x^2+(a+b)x+ab ~~&x\searrow~\swarrow a~~~~~~ax\\
&=(x+a)(x+b) ~ ~~&x\nearrow~\nwarrow b~~~~~~bx\\
\hline &&~~~~~~~~~~~~~~(a+b)x\\
\\
&acx^2+(ad+bc)x+bd ~~&ax\searrow~\swarrow b~~~~~~bcx\\
&=(ax+b)(cx+d) ~ ~~&cx\nearrow~\nwarrow d~~~~~~adx\\
\hline &&~~~~~~~~~~~~~~(ad+bc)x\\
\end{aligned}$

(IV) By grouping, splitting, or inserting terms to obtain common factors.

(V) By substituting subexpressions to simplify given expression.

(VI) Coefficient-determining method. First given the structure of the product, then determine
the unknown parameters in the product by the comparison of coefficients.
(VII) Factorization of symmetric or cyclic polynomials. (cf. Lecture 15.)

Examples
Example 1. Factorize $\left(d^2-c^2+a^2-b^2\right)^2-4(b c-d a)^2$
Solution

$$
\begin{aligned}
& \left(d^2-c^2+a^2-b^2\right)^2-4(b c-d a)^2=\left(d^2-c^2+a^2-b^2\right)^2-(2 b c-2 d a)^2 \\
& =\left(d^2-c^2+a^2-b^2-2 b c+2 d a\right)\left(d^2-c^2+a^2-b^2+2 b c-2 d a\right) \\
& =\left[(d+a)^2-(b+c)^2\right] \cdot\left[(d-a)^2-(b-c)^2\right] \\
& =(d+a-b-c)(d+a+b+c)(d+b-a-c)(d+c-a-b) .
\end{aligned}
$$

Example 2. Factorize $64 x^6-729 y^{12}$.
Solution

$$
\begin{aligned}
& 64 x^6-729 y^{12}=(2 x)^6-\left(3 y^2\right)^6=\left[(2 x)^3-\left(3 y^2\right)^3\right]\left[(2 x)^3+\left(3 y^2\right)^3\right] \\
& =\left(2 x-3 y^2\right)\left[(2 x)^2+(2 x)\left(3 y^2\right)+\left(3 y^2\right)^2\right] \\
& \quad \quad\left(2 x+3 y^2\right)\left[(2 x)^2-(2 x)\left(3 y^2\right)+\left(3 y^2\right)^2\right] \\
& =\left(2 x-3 y^2\right)\left(2 x+3 y^2\right)\left(4 x^2+6 x y^2+9 y^4\right)\left(4 x^2-6 x y^2+9 y^4\right) .
\end{aligned}
$$

Example 3. Factor each of the following expressions:

(i) $2 x^2+x-6$;
(ii) $2 x^2-10 x+8$.
Solution
$
\begin{aligned}
&x^2+x-6 ~~&x\searrow~\swarrow 3~~~~~~&3x\\
&=(x+3)(x-2) ~ ~~&x\nearrow~\nwarrow -2~~~~~~&-2x\\
\hline &&&~~~~x\\
\\
&2x^2-10x+8 ~&2x\searrow~\swarrow-2~~~~~~&-2x\\
&=(2x-2)(x-4) ~ ~~&x\nearrow~\nwarrow -4~~~~~~~&-8x\\
\hline &&~~~~~~~~~~~~~~&-10x\\\end{aligned}
$

Example 4. Factorize $2 a^3+6 a^2+6 a+18$.
Solution
$$
\begin{aligned}
& 2 a^3+6 a^2+6 a+18=2\left[\left(a^3+3 a^2+3 a+1\right)+8\right]=2\left[(a+1)^3+2^3\right] \\
& =2(a+3)\left[(a+1)^2-2(a+1)+4\right]=2(a+3)\left(a^2+3\right) .
\end{aligned}
$$

Example 5. Factorize  (i) $x^4+2 x^3+7 x^2+6 x-7$;
(ii) $x^3+9 x^2+23 x+15$.

Solution Let $y=x^2+x$. Then
(i)
$$
\begin{aligned}
& x^4+2 x^3+7 x^2+6 x-7=x^2\left(x^2+x\right)+x\left(x^2+x\right)+6\left(x^2+x\right)-7 \\
& =\left(x^2+x+6\right)\left(x^2+x\right)-7=y^2+6 y-7=(y+7)(y-1) \\
& =\left(x^2+x+7\right)\left(x^2+x-1\right) .
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& x^3+9 x^2+23 x+15=x^2(x+1)+8 x(x+1)+15(x+1) \\
& =(x+1)\left(x^2+8 x+15\right)=(x+1)(x+3)(x+5) .
\end{aligned}
$$

Example 6. Factorize  (i) $(a+1)(a+2)(a+3)(a+4)-120$;
(ii) $x^5+x+1$.
Solution
(i)
$$
\begin{aligned}
& (a+1)(a+2)(a+3)(a+4)-120 \\
& =[(a+1)(a+4)][(a+2)(a+3)]-120 \\
& =\left(a^2+5 a+4\right)\left(a^2+5 a+6\right)-120 \\
& =\left[\left(a^2+5 a+5\right)-1\right]\left[\left(a^2+5 a+5\right)+1\right]-120 \\
& =\left(a^2+5 a+5\right)^2-121=\left(a^2+5 a+5\right)^2-11^2 \\
& =\left(a^2+5 a-6\right)\left(a^2+5 a+16\right)=(a-1)(a+6)\left(a^2+5 a+16\right) .
\end{aligned}
$$
(ii)
$$
\begin{aligned}
& x^5+x+1=\left(x^5-x^2\right)+\left(x^2+x+1\right) \\
& =x^2\left(x^3-1\right)+\left(x^2+x+1\right) \\
& =x^2(x-1)\left(x^2+x+1\right)+\left(x^2+x+1\right) \\
& =\left(x^2+x+1\right)\left[x^2(x-1)+1\right]=\left(x^2+x+1\right)\left(x^3-x^2+1\right) .
\end{aligned}
$$

Example 7. Factorize $(2 y-3 z)^3+(3 z-4 x)^3+(4 x-2 y)^3$.
Solution  Let $2 y-3 z=a, 3 z-4 x=b, 4 x-2 y=c$, then $a+b+c=0$.

Hence
$$
\begin{aligned}
& (2 y-3 z)^3+(3 z-4 x)^3+(4 x-2 y)^3=a^3+b^3+c^3 \\
& =\left(a^3+b^3+c^3-3 a b c\right)+3 a b c \\
& =(a+b+c)\left(a^2+b^2+c^2-b c-c a-a b\right)+3 a b c \\
& =3 a b c=3(2 y-3 z)(3 z-4 x)(4 x-2 y) .
\end{aligned}
$$

Example 8. Factorize $(3 a+3 b-18 a b)(3 a+3 b-2)+(1-9 a b)^2$.
Solution  The given expression is symmetric in $3 a$ and $3 b$, so we use $u=$

$3 a+3 b, v=(3 a)(3 b)$ to simplify the expression. Then
$$
\begin{aligned}
& (3 a+3 b-18 a b)(3 a+3 b-2)+(1-9 a b)^2=(u-2 v)(u-2)+(1-v)^2 \\
& =u^2-2 u v-2 u+4 v+v^2-2 v+1=\left(u^2-2 u v+v^2\right)+2(v-u)+1 \\
& =(v-u)^2+2(v-u)+1=(v-u+1)^2 \\
& =(9 a b-3 a-3 b+1)^2=[(3 a-1)(3 b-1)]^2=(3 a-1)^2(3 b-1)^2 .
\end{aligned}
$$
Note: Do not stop at $(v-u+1)^2$.
Example 9. Factorize $2 x^2+7 x y-4 y^2-3 x+6 y-2$.
Solution Considering $2 x^2+7 x y-4 y^2=(2 x-y)(x+4 y)$, let
$$
2 x^2+7 x y-4 y^2-3 x+6 y-2=(2 x-y+a)(x+4 y+b) .
$$

By expending the product, it is obtained that
$$
(2 x-y+a)(x+4 y+b)=2 x^2+7 x y-4 y^2+(a+2 b) x+(4 a-b) y+a b .
$$

By the comparison of coefficients, the following system of equations is obtained
$$
\begin{aligned}
a+2 b & =-3 &(6.1)\\
4 a-b & =6, &(6.2)\\
a b & =-2 . &(6.3)
\end{aligned}
$$

Then $2 \times(6.2)+(6.1)$ yields $9 a=9$, i.e. $a=1$. From $(6.2), b=6-4 a=-2$. Since $(a, b)=(1,-2)$ satisfies $(6.3)$, and it is the unique solution, we obtain
$$
2 x^2+7 x y-4 y^2-3 x+6 y-2=(2 x-y+1)(x+4 y-2) .
$$

Example 10. Given that $x^5-5 q x+4 r$ has a factor $(x-c)^2$ for some constant $c$. Prove that $q^5=r^4$.

Solution If $c=0$, then $x^2 \mid x^5-5 q x+4 r \Rightarrow r=q=0$, the conclusion is true. When $c \neq 0$, the condition means
$$
\begin{aligned}
& x^5-5 q x+4 r=\left(x^2-2 c x+c^2\right)\left(x^3+a x^2+b x+d\right) \\
& =x^5+(a-2 c) x^4+\left(c^2+b-2 a c\right) x^3+\left(a c^2-2 b c+d\right) x^2 \\
& \quad+\left(b c^2-2 c d\right) x+c^2 d .
\end{aligned}
$$

Then the comparison of coefficients yields the following equations:
$$
a=2 c, \quad b=2 a c-c^2=3 c^2, \quad d=2 b c-a c^2=4 c^3=\frac{4 r}{c^2} \Rightarrow r=c^5 .
$$

Further,
$$
5 q=2 c d-b c^2=8 c^4-3 c^4=5 c^4 \Rightarrow q=c^4 .
$$

Thus, $q^5=c^{20}=r^4$.

Testing Questions (A)

1. Factorize the following expressions (i) $x^9+7 x^6 y^3+7 x^3 y^6+y^9$;
(ii) $4 x^2+y^2+9 z^2-6 y z+12 z x-4 x y$;
(iii) $\left(x^2-1\right)(x+3)(x+5)+16$;
(iv) $\left(2 x^2-4 x+1\right)^2-14 x^2+28 x+3$;
(v) $x^3-3 x^2+(a+2) x-2 a$;
(vi) $x^{11}+x^{10}+\cdots+x^2+x+1$.
2. Factorize the following expressions
(i) $\quad x^4-2\left(a^2+b^2\right) x^2+\left(a^2-b^2\right)^2$;
(ii) $(a b+1)(a+1)(b+1)+a b$.
3. Prove that $81^6-9 \cdot 27^7-9^{11}$ is divisible by 45 .
4. Prove that $\underbrace{33 \cdots 33}_{2 n \text { digits }}-\underbrace{66 \cdots 66}_{n \text { digits }}$ is a perfect square number.
5. Factorize the following expressions
(i) $\quad\left(x^2+x-1\right)^2+x^2+x-3=0$;
(ii) $(x-y)^3+(y-x-2)^3+8$;
(iii) $(6 x+5)^2(3 x+2)(x+1)-6$;
(iv) $\left(x^2+5 x+6\right)\left(x^2+6 x+6\right)-2 x^2$;
(v) $\quad\left(x^2-2 x\right)^3+\left(x^2-4 x+2\right)^3-8\left(x^2-3 x+1\right)^3$;
(vi) $a^3+b^3+c^3+(a+b)(b+c)(c+a)-2 a b c$.
6. Use the coefficient-determining method to factorize the following expressions.
(i) Given the expression $x^2+x y-2 y^2+8 x+a y-9$. Find possible values of the constant $a$, such that the polynomial can be factorized as product of two linear polynomials.
(ii) $x^4-x^3+4 x^2+3 x+5$.

7. Given that $y^2+3 y+6$ is a factor of the polynomial $y^4-6 y^3+m y^2+n y+36$. Find the values of constants $m$ and $n$.

Testing Questions (B)

1. Factorize $2\left(x^2+6 x+1\right)^2+5\left(x^2+1\right)\left(x^2+6 x+1\right)+2\left(x^2+1\right)^2$.

2. (CHINA/2001) If $x^2+2 x+5$ is a factor of $x^4+a x^2+b$, find the value of $a+b$.

3. Factor $(a b+c d)\left(a^2-b^2+c^2-d^2\right)+(a c+b d)\left(a^2+b^2-c^2-d^2\right)$.

4. Factorize $(a y+b x)^3+(a x+b y)^3-\left(a^3+b^3\right)\left(x^3+y^3\right)$.

5. Given that $a, b, c$ are three distinct positive integers. Prove that among the numbers
$$
a^5 b-a b^5, b^5 c-b c^5, c^5 a-c a^5,
$$
there must be one that is divisible by 8 .