(3x+2) is a factor of $3 x^{3} + 2 x^{2} - 3 x - 2.$ Hence, factories the expression $3 x^{3} + 2 x^{2} - 3 x - 2.$ completely.

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Solution

Here, $f (x) = 3 x^{3} + 2 x^{2} - 3 x - 2$
So, $3 x + 2 = 0 \Rightarrow x = - 2 / 3$
Thus, remainder $= f (- 2 / 3)$
$= 3 (- 2 / 3)^{3} + 2 (- 2 / 3)^{2} - 3 (- 2 / 3) - 2$
$= - 8 / 9 + 8 / 9 + 2 - 2$
$= 0$
Therefore, $(3 x + 2)$ is a factor of $f (x) .$
Now, performing division of polynomial $f (x)$ by $(3 x + 2)$ we get
Thus, $3 x^{3} + 2 x^{2} - 3 x - 2 = (3 x + 2) (x^{2} - 1) = (3 x + 2) (x - 1) (x + 1)$

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$3 x^{3} + 2 x^{2} + 3 x + 2$ =?

(a) $(3 x - 2) (x^{2} - 1)$

(b) $(3 x - 2) (x^{2} + 1)$

(d) $(3 x + 2) (x^{2} + 1)$

Q. Evaluate the following:

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

lim x \to 3 \frac{x^{3} - 8 x^{2} + 45}{2 x^{2} - 3 x - 9}

Q. What must be subtracted from

4 x^{3} - 3 x + 5

to get

2 x^{2} - 3 x^{3} + 5 x - 2

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(3x+2) is a factor of 3x3+2x2−3x−2. Hence, factories the expression 3x3+2x2−3x−2. completely.

Here,f(x)=3x3+2x2–3x–2So, 3x+2=0⇒x=−2/3Thus, remainder =f(−2/3)=3(−2/3)3+2(−2/3)2–3(−2/3)–2=−8/9+8/9+2–2=0Therefore, (3x+2) is a factor of f(x).Now, performing division of polynomial f(x) by (3x+2) we getThus, 3x3+2x2–3x–2=(3x+2)(x2–1)=(3x+2)(x–1)(x+1)

(3x+2) is a factor of $3 x^{3} + 2 x^{2} - 3 x - 2.$ Hence, factories the expression $3 x^{3} + 2 x^{2} - 3 x - 2.$ completely.