Question

if x^3+px^2 +qx+6 leaves remainder3 when divided by x-3 and x-2 is a factor of the given polynomial,then find the value of p and q.

Answer 1

x-3=0⇒x=3
Let f(x)=x³+px² +qx+6
=(3)³+p(3)²+q(3)+6=27+p(9)+3q+6=27+9p+3q+6=33+9p+3q
⇒33+9p+3q=3
⇒9p+3q=3-33⇒9p+3q=-30⇒9p=-30-3q⇒p=(-30-3q)/9
x-2=0⇒x=2
 f(x)=x³+px² +qx+6
=(2)³+p(2)²+q(2)+6=8+p(4)+2q+6=8+4p+2q+6=14+4p+2q
14+4p+2q=0
⇒14+4(-30-3q)/9+2q=0
⇒14(-120-12q/9)+2q=0
⇒(126-120-12q+2q)/9=0
⇒(6-10q)/9=0⇒6-10q=0⇒-10q=-6⇒q=-6/-10=3/5=0.6
p=(-30-3q)/9=[-30-3(0.6)]/9=(-30-1.8)/9=-31.8/9=-3.53
∴p=-3.53,q=0.6

Answer 2

 If the given polynomial is f(x) = x3 + px2 + qx + 6 then solution will follows as:
When f(x) is divided by x - 3 and x - 2, the remainders are 3 and 0 respectively.
∴ f(3) = 3 and f(2) = 0
⇒ (3)3 + p(3)2 + q(3) + 6 = 3 and (2)3 + p(2)2 + q(2) + 6 = 0
⇒ 27+ 9p + 3q + 6 = 3 and 8 + 4p + 2q + 6 = 0
⇒ 9p + 3q + 33 = 3 and 4p + 2q + 14 = 0
⇒ 9p + 3q = -30 and 4p + 2q = -14
⇒ 3p + q = -10 ... (1) and 2p + q = -7 .... (2)
On subtracting (1) and (2), we get
3p + q - (2p + q) = -10 -(-7)
⇒ p = -10 + 7 = -3
On putting p in (2), we get
2(-3) + q = -7
⇒ q = -7 + 6 = -1