What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?(1 point) Responses 25–√3 2 root index 3 start root 5 end root 210−−√3 2 root index 3 start root 10 end root 85–√3 8 root index 3 start root 5 end root −25–√3
The correct answer is option 2: 2√(10)−−√3.
To rewrite ∛40-√3 as a product of an integer and another irrational cube root, we can simplify it as follows:
∛40-√3 = ∛(2^3 * 5) - √3 = 2∛5 - √3
Therefore, the expression ∛40-√3 can be rewritten as 2∛5 - √3.
To rewrite the irrational cube root ∛40-√3 as a product of an integer and another irrational cube root, you can follow these steps:
Step 1: Simplify the cube root ∛40-√3 as much as possible.
Step 2: Identify the perfect cube factor(s) of 40.
Step 3: Express the remaining factor(s) as another irrational cube root.
Let's begin with step 1:
Step 1: Simplify the cube root ∛40-√3
We can't simplify the cube root of 40 any further, but we can simplify the √3 by rationalizing the denominator:
∛40-√3 = ∛40-√3 * (√3)/(√3)
= ∛40√3 - 3
= 2∛10 - 3
Now, let's move on to step 2:
Step 2: Identify the perfect cube factor(s) of 40
From the prime factorization of 40 (2 * 2 * 2 * 5), we can see that 8 (2 * 2 * 2) is the largest perfect cube factor.
Step 3: Express the remaining factor(s) as another irrational cube root.
To express the remaining factor as an irrational cube root, we divide 40 by 8:
40 / 8 = 5
Therefore, we can rewrite ∛40-√3 as 2∛10 - 3, where 2 is the integer part and ∛10 is another irrational cube root.