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For a complete method to find prime factors, it could be shown on a complete factor tree with no more than one error of division by prime factors with no more than one error - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 1

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For a complete method to find prime factors, it could be shown on a complete factor tree with no more than one error of division by prime factors with no more than o... show full transcript

Worked Solution & Example Answer:For a complete method to find prime factors, it could be shown on a complete factor tree with no more than one error of division by prime factors with no more than one error - Edexcel - GCSE Maths - Question 1 - 2022 - Paper 1

Step 1

For a complete method to find prime factors, it could be shown on a complete factor tree with no more than one error of division by prime factors with no more than one error.

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Answer

To find the prime factors of the expression x29x^2 - 9, we begin by recognizing it as a difference of squares:

x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3)

Next, we can factor the individual components if applicable. In this case, neither x3x - 3 nor x+3x + 3 can be factored further into prime factors unless specific values of xx are provided.

Step 2

For complete factorization, e.g., 2, 2, 5, 5.

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Answer

The complete factorization for the expression can be expressed in terms of its prime factors. Since x29x^2 - 9 evaluates to the product of two linear terms, we can focus on the coefficients if they were numeric. Here, the coefficients of any numerical constants would be expressed similarly if needed. The final expression retains the factors (x3)(x - 3) and (x+3)(x + 3) as prime factors of this polynomial.

Step 3

For $x^2 - 9$

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Answer

Thus, the complete factorization of x29x^2 - 9 is: x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3)

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