Q13 please do it my friends
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AjayJoseph:
I cant see the 'to find' part
in Q12
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13)Case 1
(x-2)is a factor
which means p(2)=0
therefore 8+4a+2b+6=0
Case2
(x-3)leaves remainder 3
which means q(3)=3
therefore 27+9a+3b+6=3
solving them I got
a=-3
b=-1
(x-2)is a factor
which means p(2)=0
therefore 8+4a+2b+6=0
Case2
(x-3)leaves remainder 3
which means q(3)=3
therefore 27+9a+3b+6=3
solving them I got
a=-3
b=-1
Sorry as I couldnt do the Q-12
for solving I used elimination method
how you have solved the equation in Q13
13)Case 1
(x-2)is a factor
which means p(2)=0
therefore 8+4a+2b+6=0
Case2
(x-3)leaves remainder 3
which means q(3)=3
therefore 27+9a+3b+6=3
solving them I got
a=-3
b=-1
(x-2)is a factor
which means p(2)=0
therefore 8+4a+2b+6=0
Case2
(x-3)leaves remainder 3
which means q(3)=3
therefore 27+9a+3b+6=3
solving them I got
a=-3
b=-1
27+9a+3b+6=3
which means 9+3a+b+2=1 8+4a+2b+6=0 means 4+2a+b+3=0
9-4 +3a-2a +b-b +2-3 = 1-0 which means 5+a-1=1 which means a= -3 by substituting this in any one of the equations you will get b= -1
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