Question

Find the remainder when
x³+ 3 x² + 3x + 1 when divided by 3x + 1


Determine the factors of the polynomial
2 y³+ y² – 2y – 1.

Answer 1

HEY MATE the remainder is 8/ 28....

Answer 2

Answer:

$1.\ \ \frac{8}{27} \\ \\ \\ 2.\ \ 2y + 1,\ y + 1,\ y - 1$

Step-by-step explanation:

$1. \\ \\ x^3 + 3x^2 + 3x + 1 \\ \\ = x^3 + \frac{1}{3}x^2 + \frac{8}{3}x^2 + \frac{8}{9}x + \frac{19}{9}x + \frac{19}{27} + \frac{8}{27} \\ \\ = \frac{1}{3}x^2(3x + 1) + \frac{8}{9}x(3x + 1) + \frac{19}{27}(3x + 1) + \frac{8}{27} \\ \\ = (3x + 1)(\frac{1}{3}x^2 + \frac{8}{9}x + \frac{19}{27}) + \frac{8}{27} \\ \\ = \frac{1}{27}(3x + 1)(9x^2 + 24x + 19) + \frac{8}{27}$

$\\ \\ \\ From\ this,\ we\ get\ that\ when\ x^3 + 3x^2 + 3x + 1\ is\ divided\ by\ \\ \\ 3x + 1,\ we\ get\ the\ quotient\ \frac{1}{27}(9x^2 + 24x + 19)\ and\ the \\ \\ remainder\ \bold{\frac{8}{27}}. \\ \\ \\ \therefore\ \bold{\frac{8}{27}}\ is\ the\ answer.$

$2. \\ \\ 2y^3+ y^2 - 2y - 1 \\ \\ = y^2(2y + 1) - (2y + 1) \\ \\ = (2y + 1)(y^2 - 1) \\ \\ = (2y + 1)(y + 1)(y - 1) \\ \\ \\ \therefore\ \bold{2y^3 + y^2 - 2y - 1 = (2y + 1)(y + 1)(y - 1)} \\ \\ \therefore\ The\ factors\ are\ 2y + 1,\ y + 1\ \&\ y - 1.$

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