Algebra _ Factorisation, Exercises of Mathematics

Examples of algebraic factorisation problems and their solutions. It shows how to factorize polynomials of different degrees and how to identify factors using algebraic methods. useful for students studying algebra and preparing for exams or assignments on factorisation.

Typology: Exercises

Pre 2010

Available from 03/01/2022

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Algebra
Factorisation
Factorize the following:
1. x3 4x2 + 5x 2
Solution: 1 4 + 5 2 = 0 ( x 1) is a factor
1 1 4 5 2
0 1 3 2
1 3 2 0
x2 3x + 2 = ( x 1) ( x 2)
The factors are ( x 1) ( x 2) ( x 3) = ( x 1)2 ( x 2)
Factorize the following:
2. x3 + 9x2 + 23x + 15
Solution: 1 + 9 + 23 + 15 0 ( x 1) is not a factor
1 + 23 = 9 + 15 = 24 (x 1) is a factor
1 1 9 23 15
0 1 8 15
1 8 15 0
x2 + 8x + 15 = (x + 3) (x + 5)
 The factors are ( x + 1) ( x + 3) ( x + 5)
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Algebra

Factorisation Factorize the following:

1. x^3 – 4 x^2 + 5 x – 2 Solution: 1 – 4 + 5 – 2 = 0  ( x 1) is a factor 1 1  4 5  2 0 1  3 2 1  3 2 0 x 2 - 3 x + 2 = ( x – 1) ( x – 2) The factors are ( x – 1) ( x – 2) ( x – 3) = ( x – 1) 2 ( x – 2) Factorize the following: 2. x **3

  • 9** x **2
  • 23** x + 15

Solution: 1 + 9 + 23 + 15  0  ( x 1) is not a factor

1 + 23 = 9 + 15 = 24  ( x 1) is a factor  1 1 9 23 15 0  1  8  15 1 8 15 0 x^2 + 8 x + 15 = ( x + 3) ( x + 5)

 The factors are ( x + 1) ( x + 3) ( x + 5)

3. x^3  2 x^2  5 x + 6 Solution: 1  2 5 + 6 = 0  ( x  1) is a factor 1  5  2 + 6  ( x + 1) is not a factor 1 1  2  5 6 0 1  1  6 1  1  6 0 x 2  x  6 = ( x 3) ( x + 2)

 The factors are ( x  1) ( x  3) ( x + 2)

Factorize the following:

4. 2 x^4 + 7 x^3 + x^2 – 7 x3 Solution: 2 + 7 + 1  7  3 = 0 ( x 1) is a factor 2 + 1  3 = 7  7 = 0  ( x + 1) is also a factor 1 2 7 1  7  3 0 2 9 10 3  1 2 9 10 3 0 0  2  7  3 2 7 3 0 2 x 2

  • 7 x + 3 = 2 x 2
  • 6 x + x + 3 = 2 x ( x + 3) + 1 ( x + 3) = ( x + 3) (2 x + 1)

   The factors are ( x  1) ( x + 1) ( x + 3) ( 2 x + 1)

7. m 3 - 2 m 2 - 4 m + 8 Solution:

1  2  4 + 8  0 ( m  1) is not a factor

1  4  2 + 8  ( m + 1) is not a factor

In that case try m = 2 2 1  2  4 8 0 2 0  8 1 0  4 0 m^2  4 = ( m + 2) ( m  2)

 The factors are ( m  2)^2 ( m + 2)

Factorize the following:

8. a^3 – 5 a^2 – 2 a + 24 Solution: 1  5  2 + 24  0  ( a 1) is not a factor 1  2  5 + 24  1  ( a + 1) is also not a factor Try a = 2 2 1  5  2 24 0 2  6  16 1  3  8 8  ( a  2) is not a factor Try a =  2  2 1  5  2 24 0  2 14  24 1  7 12 0 a^2 – 7 a + 12 = ( a 3) ( a 4)  The factors are ( a + 2) ( a3) ( a4)

9. x 4  5 x+ 4 Solution: x 4

  • 0 x 3
  • 5 x + 0 x + 4 1 + 0 – 5 + 0 + 4 = 0 ( x 1) is a factor 1  5 + 4 = 0 ( x + 1) is also a factor 1 1 0  5 0 4 0 1 1  4  4  1 1 1  4  4 0 0  1 0 4 1 0  4 0 x 2  4 = ( x 2) ( x + 2)  The factors are ( x  2) ( x + 2) ( x1) ( x + 1)