Question

If (x-3) and (x-1/3) are the factors of the polynomial px²+3x+r, show that p=r

Answer 1

Answer:

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Answer 2

p = r = -9/10

Given:

(x-3) and (x-1/3) are the factors of the polynomial px²+3x+r,

To find:

show that p=r

Solution:

x- 3 = 0

x = 3       [ Because x - 3 is a factor, it will be equal to zero]

x- 1/3 = 0

x = 1/3

Taking the function q(x)

q(x)/10 = px²+3x+r

Putting the values of x in equation one by one

q(3) = p.3² + 3.3 + r

0 = 9p + 9 + r    [ q(3) = 0]

r = -9p - 9 ........1

Putting value of x as 1/3 in function

q(1/3) = p. (1/3)² + 3. 1/3 + r

0 = 1/9.p + 1 + r

Putting value of r in equation:

0 = p / 9 +1 - 9p - 9

p/9 -9p -8 = 0

p/9 - 9p = 8

(p - 81p)/ 9 = 8

p-81p = 72

-80 p = 72

p = -72/80

p =-9/10

Putting value of p in eq 1

r = -9(-9/10) - 9

= 81/10 - 9

=( 81 - 90)/10

= -9/10

So, the value of both p and r is -91/10

Hence, they are equal.

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