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Question 10 If both 𝑥−2 and 𝑥−1/2 are factors of 𝑝𝑥^2+5𝑥+𝑟, show that 𝑝=𝑟. If both 𝑥−2 and 𝑥−1/2 are factors of 𝑝𝑥^2+5𝑥+𝑟, Then we can write 𝒌(𝒙−𝟐)(𝒙−𝟏/𝟐)= 𝒑𝒙^𝟐+𝟓𝒙+𝒓 Where k is any constant Now, solving our equation 𝒌(𝒙−𝟐)(𝒙−𝟏/𝟐)= 𝒑𝒙^𝟐+𝟓𝒙+𝒓 𝑘[𝑥(𝑥−1/2)−2(𝑥−1/2)]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−1/2 𝑥−2𝑥+2 ×1/2]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−1/2 𝑥−2𝑥+1]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−𝑥(1/2+2) +1]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−𝑥((1 + 2 × 2)/2) +1]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−𝑥((1 + 4)/2) +1]= 𝑝𝑥^2+5𝑥+𝑟 𝑘[𝑥^2−5/2 𝑥 +1]= 𝑝𝑥^2+5𝑥+𝑟 〖𝒌𝒙〗^𝟐−𝟓𝒌/𝟐 𝒙 +𝒌 = 𝒑𝒙^𝟐+𝟓𝒙+𝒓 Comparing x2 terms 𝒌=𝒑 Comparing constant terms 𝒌=𝒓 Thus, 𝒑=𝒓 Hence proved Comparing constant terms 𝒌=𝒓 Thus, 𝒑=𝒓 Hence proved Note: By comparing x terms, we can find value of k. And hence find value of p & r

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Davneet Singh

Davneet Singh is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

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