(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$. Find the value of $p$. (b) Solve the equation $\frac{3}{2x+1} + \frac{2}{5} = \frac{2}{3x-1}$ where $x \neq -\frac{1}{2}, \frac{1}{3}$, and $x \in \mathbb{R}$.

Leaving Cert Mathematics Algebra: (a) In the expansion of $(2x + 1

(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$ - Leaving Cert Mathematics - Question 1 - 2019

Question 1

$(a)-In-the-expansion-of-$(2x-+-1)(x^{2}-+-px-+-4)$,-where-$p-\in-\mathbb{N}$,-the-coefficient-of-$x$-is-twice-the-coefficient-of-$x^{2}$-Leaving Cert Mathematics-Question 1-2019.png$

(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$. Find the value of $p$. (b)... show full transcript

Worked Solution & Example Answer:(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$ - Leaving Cert Mathematics - Question 1 - 2019

Step 1

In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$.

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Answer

To solve for $p$ , we start by expanding the expression:

$\begin{align*} (2x + 1)(x^2 + px + 4) & = 2x(x^2) + 2x(px) + 2x(4) + 1(x^2) + 1(px) + 1(4) \\ & = 2x^3 + 2px^2 + 8x + x^2 + px + 4 \\ & = 2x^3 + (2p + 1)x^2 + (8 + p)x + 4. \end{align*}$

Next, identify the coefficients:

Coefficient of $x^2$ is: $2p + 1$ .
Coefficient of $x$ is: $8 + p$ .

According to the problem, the coefficient of $x$ is twice the coefficient of $x^2$ :

$8 + p = 2(2p + 1)$

Expanding this gives:

$8 + p = 4p + 2$

Rearranging the equation:

$8 - 2 = 4p - p$

$6 = 3p$

Thus, solving for $p$ gives:

$$p = 2.$

Step 2

Solve the equation $\frac{3}{2x+1} + \frac{2}{5} = \frac{2}{3x-1}$

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Answer

First, we eliminate the fractions by finding a common denominator. The common denominator is $5(2x + 1)(3x - 1)$ :

Multiplying every term by the common denominator results in:

$15(3x-1) + 2(2x + 1)(3x-1) = 10(2x + 1).$

Expanding each term:

$15(3x - 1) = 45x - 15,$
$2(2x + 1)(3x - 1) = 2(6x^2 - 2x + 3x - 1) = 12x^2 + 2x - 2,$
$10(2x + 1) = 20x + 10.$

Thus, we have:

$45x - 15 + 12x^2 + 2x - 2 = 20x + 10.$

Combining the terms gives:

$12x^2 + 27x - 15 = 20x + 10.$

Rearranging:

$12x^2 + 7x - 25 = 0.$

Now, applying the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-7 \pm \sqrt{{7^2 - 4(12)(-25)}}}}{2(12)}.$

Calculating the discriminant:

$7^2 - 4(12)(-25) = 49 + 1200 = 1249.$

Substituting back gives:

$x = \frac{{-7 \pm \sqrt{1249}}}{24}.$

This yields the solutions for $x$ . Thus, the final answers are:

$$x = \frac{-7 + \sqrt{1249}}{24} \text{ or } x = \frac{-7 - \sqrt{1249}}{24}, \text{ ensuring both are valid as long as } x \neq -\frac{1}{2}, \frac{1}{3}.$

(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$ - Leaving Cert Mathematics - Question 1 - 2019

Worked Solution & Example Answer:(a) In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$ - Leaving Cert Mathematics - Question 1 - 2019

In the expansion of $(2x + 1)(x^{2} + px + 4)$, where $p \in \mathbb{N}$, the coefficient of $x$ is twice the coefficient of $x^{2}$.

Solve the equation \(\frac{3}{2x+1} + \frac{2}{5} = \frac{2}{3x-1}\)

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