Factorisation using Identity
(i)
a 2 + 2ab +
b 2 ≡ (a +b)2
(ii)
a 2 − 2ab +b2 ≡ (a −b)2
(iii) a2 −
b2 ≡ (a +
b )(a −b)
(iv)
a 2 + b 2 +c 2 +
2ab + 2bc + 2ca ≡
(a + b +c)2
(v)
a 3 + b 3 ≡
(a + b )(a 2 −ab +b2 )
(vi)
a 3 − b 3 ≡
(a − b )(a 2 +ab +b2 )
(vii)
a 3 + b 3 +c 3 − 3abc ≡
(a +
b +c
)(a 2 +
b 2 +c2 − ab −bc
−ca)
Note
(a + b )2 +(a − b)2
= 2(a2 + b2 ); a4
−b4 = (a2 +b2 )(a + b )(a −b)
(a + b )2 −(a −b)2 = 4ab ; a6 − b6
= (a +b)(a − b )(a2 −ab + b2 )(a2 +ab +b2 )
Progress Check
Example 3.25
Factorise
the following:
(i)
9x2 + 12xy +
4y2 (ii)
25a2 − 10a +1 (iii) 36m2 −
49n2 (iv) x3 –x (v) x4 − 16 (vi) x2
+
4y2 + 9z2 −
4xy +12yz −6xz
(i)
9x2 + 12xy +
4y2 = (3x )2 +
2(3x )(2y ) +(2y)2 [ a 2 + 2ab + b2 = (a +b)2 ]
= (3x + 2y)2
(ii)
25a2 − 10a +1
=
(5a )2 −2(5a )(1) +12
=
(5a −1)2 [ a 2 − 2ab + b2 = (a − b)2 ]
(iii)
36m2 − 49n2. = (6m)2 -(7n)2
=
(6m + 7n)(6m −7n) [a
2 − b2 = (a + b)(a −b)]
(iv) x3 − x =
x (x2 −1)
= x(x
2 −12 )
=
x (x +1)(x
−1)
(v) x4 -16 = x4 -24 [a4
− b4 =(a2 + b2)(a
+ b)(a − b)]
= (x 2 +
22 )(x2 −22 )
= (x 2 +
4)(x + 2)(x −2)
(vi) x 2 +
4y 2 +
9z 2 −
4xy +12yz
−6xz
= (−x )2 + (2y )2 +
(3z )2 + 2(−x )(2y ) +
2(2y)(3z ) + 2(3z)(−x)
= (−x + 2y + 3z)2
or (x − 2y − 3z)2
Factorise
the following:
(i)
27x 3 +125y3
(ii)
216m 3 − 343n3
(iii)
2x4 -16xy3
(iv)
8x3 + 27y3 + 64z3 − 72xyz
(i)
27x3 +125y3
= (3x)3 +(5y)2 [ (a3
+ b3 ) = (a + b)(a2
− ab + b2 ) ]
=
(3x + 5y)((3x)2 −
(3x)(5y ) + (5y)2 )
=
(3x + 5y)(9x2 −
15xy + 25y2 )
(ii)
216m 3 − 343n3 = (6m )3 −(7n)3 [a3− b3 ) = (a − b)(a 2+ ab + b2 )]
=
(6m − 7n) ((6m )2 +
(6m)(7n ) + (7n)2 )
=
(6m −7n)(36m 2 +
42mn + 49n2 )
(iii)
2x4 − 16xy3 = 2x (x 3 − 8y3 )
= 2x (x3 −(2y)3 ) [(a3
− b3 ) = (a − b)(a2 + ab + b2)]
= 2x ((x − 2y )(x2 + (x)(2y) + (2y)2 ))
= 2x (x −2y )(x2 + 2xy +4y2
)
(iv)
8x3 + 27y3 +
64z3 −72xyz
=
(2x)3 + (3y)3 + (4z)3 − 3(2x)(3y)(4z )
=
(2x + 3y + 4z )(4x 2 +
9y 2 + 16z 2 −6xy
−
12yz − 8xz)
Thinking Corner
Check 15 divides the following
(i) 20173 + 20183
(ii) 20183 – 19733
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