Q15. If f(x) =x® +px? +qx +6 has (x-2) as a factor and leaves a remainder 3, When divided by
(x-3), find the values of p and q.
D respectively. If bisector
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Let p(x) = x3 + px2 + qx +6
(x-2) is a factor of the polynomial x3 + px2 + a x +6
p(2) = 0
p(2) = 23 + p.22 + q.2 +6 =8+4p+2q+6 =14+ 4p+ 2q = 0
7 +2 p +q = 0
q= - 7 -2p -(i)
x3 + px2 + qx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + p.32 + q .3 +6= 27+9p+3q +6 =33+9p+3q= 3
11+3p +q =1 => 3p+q =-10 => q = -10-3p -(ii)
Equating the value of q from (ii) and (i) , we have
(- 7 -2p) = (-10 - 3p)
P = -3
Substituting p = -3 in (i), we get
q = - 7 -2(-3) = -7 + 6 = -1
Thus the values of p and q are -3 and -1 respectively.
Let p(x) = x3 + px2 + qx +6
(x-2) is a factor of the polynomial x3 + px2 + a x +6
p(2) = 0
p(2) = 23 + p.22 + q.2 +6 =8+4p+2q+6 =14+ 4p+ 2q = 0
7 +2 p +q = 0
q= - 7 -2p -(i)
x3 + px2 + qx +6 when divided by (x-3) leaves remainder 3.
p(3) = 3
p(3) = 33 + p.32 + q .3 +6= 27+9p+3q +6 =33+9p+3q= 3
11+3p +q =1 => 3p+q =-10 => q = -10-3p -(ii)
Equating the value of q from (ii) and (i) , we have
(- 7 -2p) = (-10 - 3p)
P = -3
Substituting p = -3 in (i), we get
q = - 7 -2(-3) = -7 + 6 = -1
Thus the values of p and q are -3 and -1 respectively.
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