Question

only part(I,ii,vii,viii)

Answer 1

1) x^2 + 13x +30

=> x^2 +5x +6x +30

=> x(x +5) +6(x +5)

=>(x +5)(x +6)


ii) a^2 -5a -66

=> a^2 -11a +6a -66

=>a(a -11) +6 (a -11)

=> (a -11)(a +6)

vii) (a -3b)^2 -9(a -3b) +20

=> let (a -3b) = r
then equation convert
r^2 -9r +20

=> r^2 -5r -4r +20

=> r ( r -5) -4( r-5)

=> (r -5 )( r -4 )
now,
put r = a -3 b

so,
(a -3b-5 ) ( a-3b -4)


viii) 15k^2 -101kl +38l^2

=>15k^2 -95kl -6kl + 38l^2

=>5k(3k -19l) -2l( 3k -19l )

=>(5k -2l) ( 3k- 19l)

Answer 2

i) x²+13x+30=0
x²+(10+3)x+30=0
x²+10x+3x+30=0
x(x+10)+3(x+10)=0
(x+3)(x+10)=0
(x+3)=0 ,. (x+10)=0
x= -3 or -10 ans

ii) a²-5a-66=0
a²-(11-6)a-66=0
a²-11a+6a-66=0
a(a-11)+6(a-11)=0
(a+6)(a-11)=0
(a+6)=0. (a-11)=0
a= -6 or 11 ans

iii) (a-3b)²-9(a-3b)+20=0
let (a-3b)=x
so, x² -9x+20=0
x²-(4+5)x+20=0
x² - 4x-5x+20=0
x(x-4)-5(x-4)=0
(x-4)(x-5)=0
x= 4 or 5
(a-3b)=x
= 5
a=5+3b
a=5+3(5-a)
a=20-3a
a= 5
or a = 4 (using x=4)

iv) 15k²-101kl+38l²
15k²-95kl-6kl+38l²
5k(3k-19l)-2l(3k-19l)
(5k-2l)(3k-19l)