Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of $a, b$, and $x \in \mathbb{R}$: \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fact, or otherwise, answer the following two questions. (i) Solve the following equation in $x$, where $a$ and $b$ are constants. \[x^2 + (a + b)x + ab = 0\] Give each answer in terms of $a$ or $b$. (ii) Simplify \[x^2 + (a + b)x + ab \to (x + a).\]

Junior Cycle Mathematics ALGEBRA - Factorising expressions: Use factors to simplify \[ \frac

Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of $a, b$, and $x \in \mathbb{R}$: \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fact, or otherwise, answer the following two questions - Junior Cycle Mathematics - Question 14 - 2018

Question 14

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Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of $a, b$, and $x \in \mathbb{R}$: \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fac... show full transcript

Worked Solution & Example Answer:Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of $a, b$, and $x \in \mathbb{R}$: \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fact, or otherwise, answer the following two questions - Junior Cycle Mathematics - Question 14 - 2018

Step 1

Use factors to simplify

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Answer

To simplify [\frac{2n^2+n-15}{n^2-9}], we first factor both the numerator and the denominator.

Step 1: Factor the Numerator The numerator (2n^2 + n - 15) can be factored as: [2n^2 + n - 15 = (2n - 5)(n + 3)]

Step 2: Factor the Denominator The denominator (n^2 - 9) can be recognized as a difference of squares, which factors as: [n^2 - 9 = (n - 3)(n + 3)]

Step 3: Combine and Simplify Now substituting the factored forms back into the fraction, we have: [\frac{(2n - 5)(n + 3)}{(n - 3)(n + 3)}] Cancelling the common factor (n + 3): [\frac{2n - 5}{n - 3}]

Step 2

Solve the following equation in x, where a and b are constants.

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Answer

To solve [x^2 + (a + b)x + ab = 0], we can use the quadratic formula: [x = \frac{{-B \pm \sqrt{{B^2 - 4AC}}}}{2A}] In our case, (A = 1), (B = a + b), and (C = ab). Thus, the solutions are: [x = \frac{{-(a + b) \pm \sqrt{{(a + b)^2 - 4(1)(ab)}}}}{2(1)}] This simplifies to: [x = \frac{{-(a + b) \pm \sqrt{{a^2 - 2ab + b^2}}}}{2}] Hence, the two solutions for (x) can be expressed as: [x = -a \quad \text{or} \quad x = -b]

Step 3

Simplify x^2 + (a + b)x + ab -> (x + a).

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101 rated

Answer

To simplify the expression [x^2 + (a + b)x + ab], we observe that we can factor this expression appropriately. Notably, we rewrite the expression as: [(x + a)(x + b)] Thus, the simplified form of the expression is: [x^2 + (a + b)x + ab = (x + a)(x + b)]

Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of \(a, b\), and \(x \in \mathbb{R}\): \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fact, or otherwise, answer the following two questions - Junior Cycle Mathematics - Question 14 - 2018

Worked Solution & Example Answer:Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] For all values of \(a, b\), and \(x \in \mathbb{R}\): \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using this fact, or otherwise, answer the following two questions - Junior Cycle Mathematics - Question 14 - 2018

Use factors to simplify

Solve the following equation in x, where a and b are constants.

Simplify x^2 + (a + b)x + ab -> (x + a).

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