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3 years ago

What are the zeroes of P(x) = x3 – 6x2 - x +62

Mathematics

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

x = 6

x = -1

x = 1

Step-by-step explanation:

Given:

Correct equation;

P(x) = x³ - 6x² - x + 6

Computation:

x³ - 6x² - x + 6

x²(x-6)-1(x-6)

(x-6)(x²-1)

we know that;

a²-b² = (a+b)(a-b)

So,

(x-6)(x²-1)

(x-6)(x+1)(x-1)

So,

zeroes are;

x = 6

x = -1

x = 1

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What are the zeroes of P(x) = x3 – 6x2 - x +62​

What are the zeroes of P(x) = x3 – 6x2 - x +62