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Question

Factorise the following expressions.
(i) a2+8a+16
(ii) p210p+25
(iii) 25m2+30m+9
(iv) 49y2+84yz+36z2
(v) 4x28x+4
(vi) 121b288bc+16c2
(vii) (l+m)24lm
(viii) a4+2a2b2+b4

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Solution

(i) Given a2+8a+16
Here, the middle term 8a split into two terms 4a and 4a such that its product 4a×4a=16a2 is same as product of last term and first term of given equation.
=a2+4a++4a+16
=a(a+4)+4(a+4)
=(a+4)(a+4)

Similar way,
(ii) p210p+25
Here the middle term 10p=5p5p
=p25p5p+25
=p(p5)5(p5)
=(p5)(p5)

(iii) 25m2+30m+9
Here, middle term 30m=15m+15m
=25m2+15m+15m+9
=5m(5m+3)+3(5m+3)
=(5m+3)(5m+3)

(iv) 49y2+84yz+36z2
Middle term 84yz=42yz+42yz
=49y2+42yz+42yz+36z2
=7y(7y+6z)+6z(7y+6z)
=(7y+6z)(7y+6z)

(v) 4x28x+4
Middle term 8x=4x4x
=4x24x4x+4
=4x(x1)4(x1)
=(4x4)(x1)
=4(x1)(x1)

(vi) 121b288bc+16c2
Here, middle term 88bc=44bc44bc
=121b244bc44bc+16c2
=11b(11b4c)4c(11b4b)
=(11b4c)(11b4c)

(vii) (l+m)24lm
=l2+m2+2lm4lm (Since (a+b)2=a2+2ab+b2)
=l2+m22lm
It is the form of a2+b22ab=(ab)2
Replace a by l and b by m, we get
l2+m22lm=(lm)2
=(lm)(lm)


(viii) a4+2a2b2+b4
=(a2)2+2a2b2+(b2)2
It is in the form of x2+2xy+y2=(x+y)2
Replace x by a2 and y by b2, we get
(a2)2+2a2b2+(b2)2=[a2+b2]2
=(a2+b2)(a2+b2)

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