Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$. (b) Hence, using algebraic integration, find the exact value of $$\int \frac{x^2 + 8x - 3}{x + 2} \ dx$$ giving your answer in the form $a + b \ln 2$ where $a$ and $b$ are integers to be found.

A-Level Edexcel Maths Pure Applications of Differentiation: Given that $$\frac{x^2 + 8x

Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$ - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

Question 7

$Given-that--$$\frac{x^2-+-8x---3}{x-+-2}-\equiv-Ax-+-B-+-\frac{C}{x-+-2}$$-where-$x-\in-\mathbb{R},-x-\neq--2$,--find-the-values-of-the-constants-$A$,-$B$-and-$C$-Edexcel-A-Level Maths Pure-Question 7-2020-Paper 2.png$

Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$. ... show full transcript

Worked Solution & Example Answer:Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$ - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

Step 1

find the values of the constants A, B and C

96%

114 rated

Answer

To find the values of constants $A$ , $B$ , and $C$ , we will perform polynomial long division on the expression ( \frac{x^2 + 8x - 3}{x + 2} ).

Polynomial Long Division: Divide $x^2 + 8x - 3$ by $x + 2$ . The first term of the quotient is (x), since (x(x + 2) = x^2 + 2x).
Subtract and Simplify:
$(x^2 + 8x - 3) - (x^2 + 2x) = 6x - 3.$
Next Term of the Quotient: Now divide $6x$ by $x + 2$ , giving (6). Multiply and subtract:
$(6x - 3) - (6x + 12) = -15.$
Constants A, B, C: Hence, we find that: (A = 1), (B = 6), (C = -15).

Step 2

find the exact value of $ \int \frac{x^2 + 8x - 3}{x + 2} \ dx $

99%

104 rated

Answer

We have already established that:

$\frac{x^2 + 8x - 3}{x + 2} = x + 6 - \frac{15}{x + 2}$

Thus, we can rewrite our integral as follows:

$\int \left(x + 6 - \frac{15}{x + 2}\right) dx$

Integrate Each Term:
- The integral of $x$ is (\frac{x^2}{2}).
- The integral of $6$ is (6x).
- The integral of $-\frac{15}{x + 2}$ is (-15 \ln|x + 2|).

Therefore, we get:

$\int \frac{x^2 + 8x - 3}{x + 2} dx = \frac{x^2}{2} + 6x - 15 \ln|x + 2| + C.$

Evaluate at Limits: Given the integration bounds or definite limits, evaluate the expression accordingly.
Final Result: We need to express it in the form $a + b \ln 2$ . We look specifically at the $\ln$ component's behavior relative to $x + 2$ .

Thus, we can simplify:

Setting $x = 0$ gives us $C + 6(0) - 15 \ln(2) = -54 + 30 \ln(2)$ .

This leads us to find integers $a$ and $b$ such that:

$a = -54, \ b = -30.$

Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$ - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

Worked Solution & Example Answer:Given that $$\frac{x^2 + 8x - 3}{x + 2} \equiv Ax + B + \frac{C}{x + 2}$$ where $x \in \mathbb{R}, x \neq -2$, find the values of the constants $A$, $B$ and $C$ - Edexcel - A-Level Maths Pure - Question 7 - 2020 - Paper 2

find the values of the constants A, B and C

find the exact value of \( \int \frac{x^2 + 8x - 3}{x + 2} \ dx \)

Join the A-Level students using SimpleStudy...