Question

If
$x { }^{3} + px {}^{2} + qx + 6$
leaves reminder 3 when divided by x-3 and leaves zero when divided by x-2 . Find the values of p and q. ​

Answer 1

SOLUTION:-

Given:

Polynomial: f(x)=x³ + px² + qx + 6

When f(x) is divided by x-3 & x-2, the remainders are 3 & 0 respectively;

Therefore,

CASE 1:

x-3 =0

=) x= 3

f(3)= 3

=) (3)³ + p(3)² + q(3) + 6=3

=) 27 + 9p + 3q +6 =3

=) 9p + 3q + 33= 3

=) 9p + 3q = 3-33

=) 9p + 3q = -30

=) 3p + q = -10.............(1)

CASE 2:

f(2)=0

=) (2)³ + p(2)² + q(2) +6 =0

=) 8 + p(4) + 2q + 6=0

=) 8 + 4p + 2q + 6 =0

=) 4p + 2q +14 =0

=) 4p + 2q = -14

=) 2p + q = -7.................(2)

On subtracting, equation (1) & (2), we get;

=) 3p + q - (2p + q) = -10-(-7)

=) 3p + q - 2p -q = -10 +7

=) 3p - 2p = -3

=) p= -3

So,

Putting the value of p in equation (2), we get;

=) 2(-3) + q = -7

=) -6 + q = -7

=) q= -7 +6

=) q = -1

Hence,

The value of p is -3 &

The value of q is -1.

Hope it helps ☺️

Answer 2

Answer :-

Value of p is - 3 and the value of q is - 1.

Explanation :-

Let f(x) = x³ + px² + qx + 6

When f(x) is divided by (x - 3) gives remainder 3

Finding zero of (x - 3)

x - 3 = 0

x = 3

By remainder theorem, f(3) is the remainder

⇒ f(3) = 3

⇒ (3)³ + p(3)² + q(3) + 6 = 3

⇒ 27 + p(9) + 3q + 6 = 3

⇒ 27 + 9p + 3q = 3 - 6

⇒ 9p + 3q = - 3 - 27

⇒ 9p + 3q = - 30

⇒ 3(3p + q) = - 30

⇒ 3p + q = - 30/3

⇒ 3p + q = - 10

⇒ q = - 10 - 3p

When f(x) is divided by (x - 2) gives remainder 0

Finding the zero of (x - 2)

x - 2 = 0

x = 2

By Remainder theorem, f(2) is the remainder

⇒ f(2) = 0

⇒ (2)³ + p(2)² + q(2) + 6 = 0

⇒ 8 + p(4) + 2q = - 6

⇒ 8 + 4p + 2q = - 6

⇒ 4p + 2q = - 6 - 8

⇒ 4p + 2q = - 14

⇒ 2(2p + q) - 14

⇒ 2p + q = - 14/2

⇒ 2p + q = - 7

Substitutute q = - 10 - 3p in the above equation

⇒ 2p + (- 10 - 3p) = - 7

⇒ 2p - 10 - 3p = - 7

⇒ - p - 10 = - 7

⇒ - 10 + 7 = p

⇒ - 3 = p

⇒ p = - 3

Substitute p = - 3 in q = - 10 - 3p

⇒ q = - 10 - 3(-3)

⇒ q = - 10 + 9

⇒ q = - 1

∴ the value of p is - 3 and the value of q is - 1.