Question

verify whether the following are zero of the polynomial indicated them :- p (x)= kx square - 3 x + k , x=-1​

Answer 1

Answer:

here is your answer .........

Answer 2

$x = \: - 1 \\ \\ p(x) = k {x}^{2} - 3x + k \\ \\ p( - 1) =0\\ \\k \times( { - 1})^{2} - 3( - 1) + k = 0\\ \\ k + 3 + k = 0 ......(i)\\ \\ 2k + 3 = 0 \\ \\ k = \frac{ - 3}{2} \\ \\ from \: eq.(i) \frac{( -3 )}{2} + 3 + \frac{(-3)}{2} \\ \\ = \frac{( - 6)}{2} + 3 \\ \\ = \frac{ (- 6) + 6}{3} \\ \\ = \frac{0}{3} = 0$

Hence x= -1 is zero of p(x)

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