Question

The value of p(-2/3) for p(y) = 2y^3 - y^2 - 13y - 6 is:

A) 4  B) -4  C) -44/27  D) 44/27

I NEED WITH STEPS PLEASE! EXTREME URGENT!!! NEED IT NOW PLEASE!!!

Answer 1

p(y) = 2y³ - y² -13y - 6
for finding p(-2/3) put y = -2/3 in the given polynomial
p(-2/3) = 2(-2/3)³ - (-2/3)² - 13(-2/3) - 6
p(-2/3) = -16/27 - 4/9 + 26/3 - 6
p(-2/3) = (-16 - 4*3 + 26*9 - 6*27)/27
p(-2/3) = (-16 - 12 + 234 - 162)/7
p(-2/3) = 44/27

Answer 2

Answer:

Option D

$p(\frac{-2}{3})$ = 44/27

Step-by-step explanation:

$p(y) = 2y^3 - y^2 - 13y - 6$

For finding $p(\frac{-2}{3})$, we have to substitute y in the first equation with $\frac{-2}{3}$.

⇒$p(\frac{-2}{3}) = 2 \times (\frac{-2}{3})^3 -(\frac{-2}{3})^2 - 13 \times (\frac{-2}{3}) - 6$

              $= 2 \times (\frac{-8}{27}) -(\frac{4}{9}) + (\frac{26}{3}) - 6$

                $= \frac{-16}{27} -\frac{4}{9} + \frac{26}{3}- 6$

               $= \frac{-16}{27} -\frac{12}{27} + \frac{234}{27}- 6$

               $= \frac{234 - 16 - 12}{27}- 6$

               $= \frac{206}{27}- \frac{162}{27}$

              $= \frac{206-162}{27}$

  $p(\frac{-2}{3})$   $= \frac{44}{27}$

Thus the answer is option D : 44/27

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