Question

If one of the zeroes of the polynomial p(y)ax^2-3(a-1)x-1 then find the value of a

Answer 1

Step-by-step explanation:

p(x) = ax²- 3(a-1)x -1

Given, 1 is its zero

p(x) = a(1)² -3(a-1)1 -1 = 0

a - 3a+3 -1 = 0

-2a + 2 = 0

-2a = -2

a = -2/-2

a = 1

Answer 2

${\red{\huge{\underline{\mathbb{Answer:-}}}}}$

$Given \\ 1 \: is \: a \: zero \: of \: polynomial \: \\ \\ by \: remainder \: theorem \: \\ p(x) = a {x}^{2} - 3(a - 1)x - 1 \\ \\ p(1) = 0 \\ \\ a {(1)}^{2} - 3(a - 1)1 - 1 = 0 \\ \\ {a}^{2} - 3a + 3 - 1 = 0 \\ \\ - 2a + 2 = 0 \\ \\ 2a = 2 \\ \\ a = 1 \\ \\ the \: value \: of \: a \: is \: 1.$

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