Question

Given that (3x+2) is a factor of 3x^3+bx^2-3x-2. find the value of b?

Answer 1

`b = 2`.

Explanation:

By the factor theorem, if a monomial of the form
`(x - a)` is a factor of the polynomial
`(3\, x^(3) + b\, x^(2) - 3\, x - 2)`, then substituting in
`x = a` would set this polynomial to
`0`.

Notice that in the factor theorem, the coefficient of
`x` in
`(x - a)` is
`1`, and the sign in front of the constant
`a` is a minus sign. Rewrite the given factor
`(3\, x + 2)` in this form to find the value of
`a\!`:

`\begin{aligned}& (3\, x + 2) \\ =\; & 3\, (x + (2/3)) \\ =\; & 3\, (x - \underbrace{(-2/3)}_(a)) \end{aligned}`.

Thus,
`x = (-2/3)` should set the polynomial
`(3\, x^(3) + b\, x^(2) - 3\, x - 2)` to
`0`:

`\begin{aligned} & (3\, x^(3) + b\, x^(2) - 3\, x - 2) \\ =\; & 3* (-2/3)^(3) + b* (-2/3)^(2) - 3 * (-2/3) - 2 \\ =\; & -3* (2/3)^(3) + b* (2/3)^(2) + 3 * (2/3) - 2 \\ =\; & -(8)/(9) + (4\, b)/(9) + 2 - 2 \\ =\; & (-8 + 4\, b)/(9)\end{aligned}`.

Set this expression to
`0` and solve for
`b`:

`\displaystyle (-8 + 4\, b)/(9) = 0`.

`b = 2`.