if one zero of the quadratic polynomial (x-1)(x^2+kx+1) is -3 then value of k is

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Solution

$Let p (x) = (x - 1) (x^{2} + k x + 1) since - 3 is a zero of p (x), then p (- 3) = 0 \Rightarrow (- 3 - 1) [{(- 3)}^{2} + k (- 3) + 1] = 0 \Rightarrow - 4 (9 - 3 k + 1) = 0 \Rightarrow - 4 (10 - 3 k) = 0 \Rightarrow - 40 + 12 k = 0 \Rightarrow 12 k = 40 \Rightarrow k = \frac{40}{12} = \frac{10}{3}$

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(k - 1) x^{2} + k x + 1

(k - 1) x^{2} + k x + 1

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(B) \frac{- 4}{3}

(C) \frac{2}{3}

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If one of the zeros of quadratic polynomial (k - 1) x^{2} + k x + 1 is - 3,

then find the value of k .

(k - 1) x^{2} + k x + 1

α

f (x) = a x^{2} + b x + c t h e n, f (α)

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