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3 years ago

Factor the polynomial F(x) =x^3-x^2-4x+4 completely.

Mathematics

1 answer:

Norma-Jean [14]3 years ago

6 0

Answer:

Step-by-step explanation:

PART 1:

Possible roots by noticing the coefficient of first term:

x = 1, -1

PART 2:

1 | 1 -1 -4 4

1 0 -4

1 0 -4 0

The remainder is zero, hence one factor is (x-1)

PART 3:

$x^3-x^2-4x+4=(x^2-4)(x-1)\\\\x^3-x^2-4x+4=(x+2)(x-2)(x-1)$

PART 4:

$f(x)=(x+2)(x-2)(x-1)\\\\=(x^2-4)(x-1)\\\\=x^3-4x-x^2+4\\\\=x^3-x^2-4x+4$

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Answer:

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Step-by-step explanation:

An equation in the vertex form is written as

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Then the vertex is:

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Therefore The encuacion excrita in the form of vertice is:

$y = a(x +\frac{1}{2}) ^ 2 -\frac{13}{4}$

To find the coefficient a we substitute a point that belongs to the function $y = 9x^2 + 9x - 1$

The point (0, -1) belongs to the function. Thus.

$-1 = a(0 + \frac{1}{2}) ^ 2 -\frac{13}{4}$

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Then the written function in the form of vertice is

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