FACTORING POLYNOMIALS EXAMPLES

Question 1 :

Factorise the following expressions:

(i) 2a² + 4a² b + 8a² c

Solution :

  =  2a² + 4a² b + 8a² c

By factoring 2a from the three terms, we get

  =  2a (a + 2ab + 4ac)

(ii)  ab - ac - mb + mc

Solution :

  =  ab - ac - mb + mc

  =  a(b - c) - m(b - c)

  =  (a - m) (b - c)

Question 2 :

Factorise the following:

(i) x² + 4x + 4

Solution :

  =  x² + 4x + 4

  =  x² + 2 ⋅ x ⋅ 2 + 2²

  =  (x + 2)²

(ii)  3a² - 24ab + 48b²

Solution :

  =  3a² - 24ab + 48b²

  =  3 (a² - 8ab + 16b²)

  =  3 [a² - 2a(4b) + (4b)²]

  =  3 (a - 4b)²

(iii)  x⁵ - 16x

Solution :

  =  x⁵ - 16x

=  x(x⁴ - 16)

=  x [(x²)² - 4²]

=  x(x² + 4)(x² - 4)

=  x(x² + 4)(x + 2)(x - 2)

(iv)  m² + 1/m² - 23

Solution :

m² + 1/m² - 23  =  m² + 1/m² - 25 + 2

  =  (m² + 1/m²  + 2) - 25

  =  (m + (1/m))² - 5²

  =  (m + (1/m) + 5) (m + (1/m) - 5)

(v)  6 - 216 x²

Solution :

  =  6 - 216 x²

  =  6(1 - 36x²)

  =  6[1 - (6x)²]

=  6(1 + 6x)(1 - 6x)

(vi)  a² + 1/a² - 18

Solution :

a² + 1/a² - 18  =  a² + 1/a² - 16 - 2

  =  (a² + 1/a² - 2) - 16

  =  (a - (1/a))² - 4²

  =  (a + (1/a) + 4) (a + (1/a) - 4)

Question 3 :

Factorise the following:

(i)  4x² +  9y² + 25z² + 12xy + 30yz + 20xz

Solution :

  =  4x² +  9y² + 25z² + 12xy + 30yz + 20xz

  =  (2x)²+(3y)²+(5z)²+2(2x)(3y)+2(3y) (5z)+2(5z)(2x)

  =  (2x + 3y + 5z)²

(ii)  25x² + 4y² + 9z² - 20xy + 12yz - 30zx

Solution :

  =  25x² + 4y² + 9z² - 20xy + 12yz - 30zx

  =  (5x)²+(-2y)²+(-3z)²+2(5x)(-2y)+2(-2y)(3z)+2(-3z)(5x)

  =  (5x - 2y - 3z)²

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