On dividing x3-3 x2+x+2 by a polynomial a cos. the quotient and remainder were x-2 and -2 x+4 respectively. Find g(x).
OR
If the sum of the squares of zeros of the quadratic polynomial f(x)=x2-8 x+k is 40, find the value of k
We know that,
p(x) = g(x)q(x) + r(x)
Putting values, we get
x3 – 3x2 + x + 2 = g(x)(x – 2) – 2x + 4
⇒ g(x)(x – 2) = x3 – 3x2 + x + 2 + 2x – 4
⇒ g(x)(x – 2) = x3 – 3x2 + 3x – 2
⇒ g(x)(x – 2) = x3 – 2x2 – x2 + 2x + x – 2
⇒ g(x)(x – 2) = x2(x – 2) – x(x – 2) + x – 2
⇒ g(x)(x – 2) = (x – 2)(x2 – x + 1)
⇒ g(x) = x2 – x + 1
OR
Let zeroes of x2 – 8x + k = 40 be α and β
⇒ α + β = 8 [-b/a]
⇒ αβ = k [c/a]
Given,
α2 + β2 = 40
⇒ (α + β)2 – 2αβ = 40
⇒ 82 – 2k = 40
⇒ 64 – 2k = 40
⇒ 2k = 24
⇒ k = 12
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