Question

4-4 practice
adding and subtracting rational expressions
find the lcm for each group of expressions.

1. $2x^2 - 8x + 8$ and $3x^2 + 27x - 30$

$2x^2 - 8x + 8 = 2(x^2 - 4x + 4) = 2(x - 2)^2$
$3x^2 + 27x - 30 = 3(x^2 + 9x - 10) = 3(x + 10)(x - 1)$
$2 \times 3 \times (x - 2)^2 \times (x + 10) \times (x - 1)$

1. $4x^2 + 12x + 9$ and $4x^2 - 9$

Explanation:

Step1: Factor the first expression

Factor \(2x^{2}-8x + 8\). First, factor out the greatest common factor (GCF) of 2:
\(2x^{2}-8x + 8=2(x^{2}-4x + 4)\)
Then, factor the quadratic \(x^{2}-4x + 4\) which is a perfect square trinomial: \(x^{2}-4x + 4=(x - 2)^{2}\)
So, \(2x^{2}-8x + 8 = 2(x - 2)^{2}\)

Step2: Factor the second expression

Factor \(3x^{2}+27x-30\). First, factor out the GCF of 3:
\(3x^{2}+27x-30 = 3(x^{2}+9x - 10)\)
Then, factor the quadratic \(x^{2}+9x - 10\). We need two numbers that multiply to - 10 and add to 9. The numbers are 10 and - 1:
\(x^{2}+9x - 10=(x + 10)(x - 1)\)
So, \(3x^{2}+27x-30=3(x + 10)(x - 1)\)

Step3: Find the LCM

To find the LCM of the two factored expressions \(2(x - 2)^{2}\) and \(3(x + 10)(x - 1)\), we take the highest power of each distinct factor that appears in the factorizations.
The distinct factors are 2, 3, \((x - 2)^{2}\), \((x + 10)\), and \((x - 1)\).
Multiply these factors together:
\(LCM=2\times3\times(x - 2)^{2}\times(x + 10)\times(x - 1)=6(x - 2)^{2}(x + 10)(x - 1)\)

Answer:

\(6(x - 2)^{2}(x + 10)(x - 1)\)