Photo AI
(a) Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] (b) 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
![(a)-Use-factors-to-simplify--\[-\frac{2n^2+n-15}{n^2-9}-\]--(b)-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.png](https://cdn.simplestudy.io/assets/backend/uploads/question/20230310111035.png)
(a) Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] (b) For all values of $a$, $b$, and $x \in \mathbb{R}$: \[(x + a)(x + b) = x^2 + (a + b)x + ab.\] Using... show full transcript
Worked Solution & Example Answer:(a) Use factors to simplify \[ \frac{2n^2+n-15}{n^2-9} \] (b) 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 \[\frac{2n^2+n-15}{n^2-9}\]
Answer
To simplify [\frac{2n^2+n-15}{n^2-9}], we first factor both the numerator and the denominator:
-
Factor the numerator: We look for two numbers that multiply to (-30) (the product of (2 \cdot -15)) and add to (1) (the coefficient of (n)). These numbers are (6) and (-5), so we can rewrite the numerator as: [2n^2 + 6n - 5n - 15 = 2n(n + 3) - 5(n + 3) = (2n - 5)(n + 3).]
-
Factor the denominator: The expression (n^2 - 9) is a difference of squares, so we can factor it as: [n^2 - 9 = (n - 3)(n + 3).]
-
Combine the factored forms: Now our original expression becomes: [\frac{(2n - 5)(n + 3)}{(n - 3)(n + 3)}.]
-
Cancel common factors: The ((n + 3)) in the numerator and denominator cancels out (as long as (n \neq -3)): [\frac{2n - 5}{n - 3}.]
Thus, the simplified form is [\frac{2n - 5}{n - 3}.]
Step 2
Solve the following equation in $x$, where $a$ and $b$ are constants: \[x^2 + (a + b)x + ab = 0\]
Answer
To solve this quadratic equation, we can use the quadratic formula:[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, ] where (A = 1), (B = (a + b)), and (C = ab). Applying the formula:
-
Find : [B^2 - 4AC = (a + b)^2 - 4(1)(ab) = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a - b)^2.]
-
Use the quadratic formula: [x = \frac{-(a + b) \pm \sqrt{(a - b)^2}}{2} = \frac{-(a + b) \pm (a - b)}{2}.]
-
Find the two solutions: The two roots are:
- First root: [x_1 = \frac{-2b}{2} = -b.]
- Second root: [x_2 = \frac{-2a}{2} = -a.]
Thus, the solutions are [x = -a \text{ or } x = -b.]
Step 3
Simplify \[x^2 + (a + b)x + ab \div (x + a).\]
Answer
To simplify the expression [x^2 + (a + b)x + ab \div (x + a),] we can perform polynomial long division:
-
Divide the leading term: The leading term (x^2) divided by (x) gives us (x).
-
Multiply and subtract: Multiply (x + a) by (x): [(x)(x + a) = x^2 + ax.] Subtract this from the original polynomial: [(x^2 + (a + b)x + ab) - (x^2 + ax) = (b)x + ab.]
-
Repeat for the next term: Now, divide (bx) by (x + a). This gives us (b). Multiply and subtract: [(b)(x + a) = bx + ab.] So, now we have: [(b)x + ab - (bx + ab) = 0.]
This shows that the remainder is zero, proving that the expression simplifies to [x + b.]