If (k+ y) is a factor of each of the polynomials y² + 2y -15 and y³ + a , find values of k and a.
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Answered by
53
Let p(x) = y² +2y -15………(1)
Since(k+y) is a factor of p(x) ,
Put y = -k in eq 1
(-k)² +2(-k) -15 =0
k² -2k -15=0
k² - 5k +3k -15=0
k(k -5) +3(k-5)= 0
(k+3) or (k-5)= 0
k= -3 or k = 5
Let q(x) = y³+a
Put y = -k in q(x)
(-k)³+a= 0
-k³ +a= 0………….(2)
Put k = -3 in eq 2
-(-3)³ +a= 0
-(-27)+a = 0
27 +a = 0
a = -27
Put k = 5 in eq 2
-(5)³ +a= 0
-(125)+a = 0
-125 +a = 0
a = 125
Hence, the value of k is -3 and 5 and value of a is -27 and 125.
HOPE THIS WILL HELP YOU...
Since(k+y) is a factor of p(x) ,
Put y = -k in eq 1
(-k)² +2(-k) -15 =0
k² -2k -15=0
k² - 5k +3k -15=0
k(k -5) +3(k-5)= 0
(k+3) or (k-5)= 0
k= -3 or k = 5
Let q(x) = y³+a
Put y = -k in q(x)
(-k)³+a= 0
-k³ +a= 0………….(2)
Put k = -3 in eq 2
-(-3)³ +a= 0
-(-27)+a = 0
27 +a = 0
a = -27
Put k = 5 in eq 2
-(5)³ +a= 0
-(125)+a = 0
-125 +a = 0
a = 125
Hence, the value of k is -3 and 5 and value of a is -27 and 125.
HOPE THIS WILL HELP YOU...
Answered by
8
Since Y=-k is factor
then k^2 -2k-15=0
k^2-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=0
k=-3,+5
now it is factor of second one ,,, hence
-k^3+a =0
a= k^3= (-3)^3 & (+5)^3
a = -27 & 125
then k^2 -2k-15=0
k^2-5k+3k-15=0
k(k-5)+3(k-5)=0
(k+3)(k-5)=0
k=-3,+5
now it is factor of second one ,,, hence
-k^3+a =0
a= k^3= (-3)^3 & (+5)^3
a = -27 & 125
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