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Anna71 [15]
2 years ago
15

Helppp quick enough points

Mathematics
1 answer:
Scilla [17]2 years ago
7 0

2a)

To verify, set the factors to zero.

(x - 2) = 0, x = 2

(x + 3) = 0, x = -3

Insert zero's in the equation:

<u>when x = 2</u>

3(2)³ + 2(2)² - 19(2) + 6 = 0

<u>when x = -3</u>

3(-3)³ + 2(-3)² - 19(-3) + 6 = 0

Hence verified the factors.

2b)

= \sf \dfrac{\left(3x^3+2x^2-19x+6\right)}{\left(x-2\right)\left(x+3\right)}

\sf = \dfrac{\left(3x-1\right)\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}

\sf = 3x -1

The remaining factor is (3x - 1)

2c)

Set the factors to zero to find real zeros of f

  • (3x - 1) = 0, x = 1/3
  • (x - 2) = 0, x = 2
  • (x + 3) = 0, x = -3

2d)

<u>Complete factorization</u>: (3x - 1)(x - 2)(x + 3)

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