If both x - 2 and x−1/2 are factors of px^2 + 5x + r, then p = r p + r = 0 2p + r = 0 p + 2r = 0

Summer Bush

Summer Bush

Answered question

2023-03-20

If both x - 2 and x−1/2 are factors of px^2 + 5x + r, then
p = r
p + r = 0
2p + r = 0
p + 2r = 0

Answer & Explanation

sazasukerqa7r

sazasukerqa7r

Beginner2023-03-21Added 4 answers

p = r
(x-2) and (x−12) are the factors of p(x) = px^2+5x+r
Let x - 2 = 0, then x = 2
∴f(2)=p(2)2+5× 2 + r = 4p + 10 + r = 4p + r + 10
∴ x - 2 is a factor of f(x)
∴ x - 2 is a factor of f(x) ∴ Remainder = 0
∴ 4p + r + 10 = 0 ⇒ 4p + r =-10 ...........(i)
Again let x - 12 = 0, then x = 12
∴f(12)=p(12)2+5×12+r
=p4+52 + r
∴x−12 is a factor of f(x) ∴ Remainder = 0
∴p4+52+r=0⇒p4+r=−52
⇒ p + 4r =- 10 .........(ii)
Subtracting (i) from (ii) - 3p + 3r = 0 ⇒ 3p = 3r
⇒ p = r

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