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Sagot :
SOLUTION
STEP1
To write 63 as a product of a perfect-square factor and a factor that does not contain a perfect square, we would need to find all the prime factors.
The number 63 is a composite number. Now let us find the prime factors associated with 63.
The first step is to divide the number 63 with the smallest prime factor,i.e. 2.
63 ÷ 2 = 31.5; fraction cannot be a factor. Therefore, moving to the next prime number
Divide 63 by 3.
63 ÷ 3 = 21
Again divide 21 by 3 and keep on diving the output by 3 till you get 1 or a fraction.
21 ÷ 3 = 7
7 ÷ 3 = 2.33; cannot be a factor. Now move to the next prime number 7.
Dividing 7 by 7 we get,
7 ÷ 7 = 1
We have received 1 at the end and it doesn’t have any factor. Therefore, we cannot proceed further with the division method. the prime factorization of 63 is 3 × 3 × 7
Therefore, the answer to step 1 is
[tex]9\times7[/tex]STEP 2
Rewrite the radicand as a product using the largest perfect square factor. This give
[tex]\begin{gathered} \sqrt[]{63}=\sqrt[]{9}\times\sqrt[]{7} \\ 9\text{ is the }largestperfectsquarefactor\text{ gotten from the previous step.} \end{gathered}[/tex]STEP 3
The perfect square is 9. The square root of the perfect square is:
[tex]\begin{gathered} \sqrt[]{9}=\sqrt[]{3\times3} \\ =3 \end{gathered}[/tex]
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