Question

if x-3 and x- 1/3 are factors of the polynomial px power 2 +3x + r ,show that p=r​

Answer 1

Answer:

If x-3 and x-1/3 are factors of px^2+3x+r, how do you show that p=r?

This question requires that you find the values of p and r, which i assume before solving, are equal.

Now, if x-3 and x-1/3 are both factors to this expression , it means their division gives a value of zero.

Now lets use x-3

>>> let x-3=0.

>>> x=3

Now substitute x=3 into the expression

>>>> Px^2+3x+r

>>> P (3)^2 + 3(3) +r = 0

>>> 9p+ r + 9= 0

>>> 9p+r= -9 ……..(1) simultaneous eqn

Now use x-1/3.

Let x-1/3=0

>>> x=1/3

Now substitute x=1/3 into the polynomial

>>> P(1/3)^2+ 3(1/3) + r=0

>>> P(1/9)+ 1 +r =0

>>> P(1/9) + r= -1

>>> P + 9r = -9…(2) simultaneous eqn

Subtract (1) from (2)

>>> 9p+ r =-9

-

P + 9r =-9

>>> 8p - 8r = 0

>>> 8p = 8r

>>> p = r

As required.

Step-by-step explanation:

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Answer 2

If (x-3) and (x-1/3) are factors of px^2+3x+r

Then,

⇒ x - 3 = 0

⇒ x = 3

Putting x = 3 is px² + 3x + r

⇒ p(3)² + 3(3) + r = 0

⇒ 9p + 9 + r = 0 --------- (1)

and

⇒ x - 1/3 = 0

⇒ x = 1/3

Putting x = 1/3 in px² + 3x + r

⇒ p(1/3)² + 3(1/3) + r = 0

⇒ (p/9) + 1 + r = 0

⇒ p + 9 + 9r = 0 ------------ (2)

Subtracting eq. (2) from (1)

⇒ (9p + 9 + r) - (p + 9 + 9r) = 0

⇒ 9p + 9 + r - p - 9 - 9r = 0

⇒ 8p - 8r = 0

⇒ p - r = 0

⇒ p = r

Henc Proved

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