Factoring Polynomials by Grouping

Algorithm for Factoring by Grouping

About factorizing a polynomial – see §19 of this Guide.

Examples of Problem Solving

Example 1. Factorize the polynomial:

a) $9x^2-27xy+7xz-21yz = (9x^2-27xy)+(7xz-21yz) = $

$ = 9x(x-3y)+7z(x-3y) = (9x+7z)(x-3y) $

b) ax+3-3x-a = (ax-3x)+(3-a) = x(a-3)-(a-3) = (x-1)(a-3)

c) $ x^3+x^2+x+1 = (x^3+x^2 )+(x+1) = x^2 (x+1)+(x+1) = (x^2+1)(x+1)$

d) $3a^2-3ax-7a+7x = (3a^2-3ax)-(7a-7x) = 3a(a-x)-7(a-x) = $

$ = (3a-7)(a-x)$

Example 2. Evaluate the expression:

a) $22,5 \cdot 11,3+77,5\cdot16,5+22,5\cdot16,5+77,5\cdot11,3 =$

$ = (22,5\cdot11,3+77,5\cdot11,3)+(22,5\cdot16,5+77,5\cdot16,5) = $

$ = 11,3(22,5+77,5)+16,5(22,5+77,5) = (11,3+16,5)(22,5+77,5) = $

$ = 27,8\cdot100 = 2780$

b) $ 38,7^2-38,7\cdot28,7+61,3^2-61,3\cdot51,3 =$

$ = (38,7^2-38,7\cdot28,7)+(61,3^2-61,3\cdot51,3) = $

$= 38,7(38,7-28,7)+61,3(61,3-51,3) = (38,7+61,3)(61,3-51,3) =$

$ = 100 \cdot 10 = 1000$

Example 3. Solve the equation:

a) x(x-3)+2(x-3) = 0

(x+2)(x-3) = 0 $\Rightarrow \left[ \begin{array}{cc} x+2 = 0 \\ x-3 = 0 \end{array} \right. \Rightarrow \left[ \begin{array}{cc} x_1 = -2 \\ x_2 = 3 \end{array} \right. $

b) $ x^3-5x^2+x-5 = 0 $

$x^2 (x-5)+(x-5) = 0 $

$ (x^2+1)(x-5) = 0 \Rightarrow \left[ \begin{array}{cc} x^2+1 = 0 \\ x-5 = 0 \end{array} \right. \Rightarrow \left[ \begin{array}{cc} no roots \\ x = 5 \end{array} \right. \Rightarrow x = 5 $

$(x^2+1 \ge 1,$ cannot be equal to 0)

c) $x^4-4x^3+x = 4$

$(x^4-4x^3 )+(x-4) = 0$

$x^3 (x-4)+(x-4) = 0$

$(x^3+1)(x-4) = 0 \Rightarrow \left[ \begin{array}{cc} x^3+1 = 0 \\ x-4 = 0 \end{array} \right. \Rightarrow \left[ \begin{array}{cc} x_1 = -1 \\ x_2 = 4 \end{array} \right. $

г*) $ x^4+x^3+2x^2+x^2+3x+2 = 0$

Note that $(x+1)(x+2) = x^2+2x+x+2 = x^2+3x+2$

Therefore:

$x^2 (x^2+3x+2)+(x^2+3x+2) = 0$

$(x^2+1)(x^2+3x+2) = 0 \Rightarrow \left[ \begin{array}{cc} x^2+1 = 0 \\ x^2+3x+2 = 0 \end{array} \right. \Rightarrow \left[ \begin{array}{cc} no roots \\ (x+1)(x+2) = 0 \end{array} \right. \Rightarrow$

$\Rightarrow \left[ \begin{array}{cc} x+1 = 0 \\ x+2 = 0\end{array} \right. \Rightarrow \left[ \begin{array}{cc} x_1 = -1 \\ x_2 = -2\end{array} \right. $

Example 4*. Factorize the trinomial:

a) $x^2+5x+6 = (x^2+3x)+(2x+6) = x(x+3)+2(x+3) = (x+2)(x+3)$

b) $x^2+7x+6 = (x^2+6x)+(x+6) = x(x+6)+(x+6) = (x+1)(x+6)$

c) $x^2+7x-30 = (x^2+10x)-(3x+30) = x(x+10)-3(x+10) = (x-3)(x+10)$

d) $x^2-5x-50 = (x^2-10x)+(5x-50) = x(x-10)+5(x-10) = (x+5)(x-10)$

Example 5. Prove that

a) $5^5-5^4+5^3-5^2$ is divisible by 13

Factorize the polynomial:

$(5^5-5^4 )+(5^3-5^2 ) = 5^4 (5-1)+5^2 (5-1) = (5^4+5^2 )(5-1) = (5^4+5^2 )\cdot4 =$

$= 4\cdot5^2 (5^2+1) = 4\cdot25\cdot26 = 4\cdot25\cdot2\cdot13 $

Q.E.D.

b) $2^{13}-2^{10}-2^9+2^6$ is divisible by 17

Factorize the polynomial:

$(2^{13}-2^{10} )-(2^9-2^6 ) = 2^{10} (2^3-1)-2^6 (2^3-1) = (2^{10}-2^6 )(2^3-1) = 2^6 (2^4-1)(2^3-1) = 64\cdot17\cdot7$

Q.E.D.

Example 6. Write as a product of three factors:

a) $ a^2 b-a^2 c-abx+acx = (a^2 b-a^2 c)-(abx-acx) = a^2 (b-c)-ax(b-c) = $

$= (a^2-ax)(b-c) = a(a-x)(b-c) $

b) $6x^2 y-10xy^2+9x^2 z-15xyz = (6x^2 y+9x^2 z)-(10xy^2+15xyz) =$

$= 3x^2 (2y+3z)-5xy(2y+3z) = (3x^2-5xy)(2y+3z) = x(3x-5y)(2y+3z)$

Example 7*. Factorize the expression:

a) $3x^{n+1}-x^n-3x+1 = (3x^{n+1}-x^n )-(3x-1) = x^n (3x-1)-(3x-1) =$

$ = (x^n-1)(3x-1) $

b) $ax^{n-1}+3x^n-3x-a = (3x^n+ax^{n-1})-(3x+a) = x^{n-1} (3x+a)-(3x+a) =$

$ = (x^{n-1}-1)(3x+a)$