Asked by Lelentle diamond sekatane
Factorise k^2-24 completely
Answers
Answered by
GPT 3.5
To factorise k^2 - 24 completely, we need to find two numbers that multiply to -24 and add up to 0 (since there is no linear term in the expression).
The two numbers that fit these criteria are -6 and 4:
-6 * 4 = -24
-6 + 4 = -2
Therefore, we can rewrite k^2 - 24 as:
k^2 - 24 = (k - 6)(k + 4)
So, the complete factorisation of k^2 - 24 is (k - 6)(k + 4).
The two numbers that fit these criteria are -6 and 4:
-6 * 4 = -24
-6 + 4 = -2
Therefore, we can rewrite k^2 - 24 as:
k^2 - 24 = (k - 6)(k + 4)
So, the complete factorisation of k^2 - 24 is (k - 6)(k + 4).
Answered by
GPT 3.5
Apologies for the confusion. Let me correct that:
To factorise k^2 - 24 completely, we can use the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b).
In this case, we have k^2 - 24, which can be rewritten as k^2 - 6^2.
Using the difference of squares formula, we get:
k^2 - 6^2 = (k + 6)(k - 6)
Therefore, the complete factorisation of k^2 - 24 is (k + 6)(k - 6).
To factorise k^2 - 24 completely, we can use the difference of squares formula, which states that a^2 - b^2 = (a + b)(a - b).
In this case, we have k^2 - 24, which can be rewritten as k^2 - 6^2.
Using the difference of squares formula, we get:
k^2 - 6^2 = (k + 6)(k - 6)
Therefore, the complete factorisation of k^2 - 24 is (k + 6)(k - 6).
Answered by
GPT 3.5
My apologies for the mistake. Let me correct that:
To factorise k^2 - 24 completely, we need to rewrite 24 as a square number. Since 24 is not a perfect square, we can write it as 6^2:
k^2 - 24 = k^2 - 6^2
Now we can apply the difference of squares formula:
k^2 - 6^2 = (k + 6)(k - 6)
Therefore, the correct factorisation of k^2 - 24 is (k + 6)(k - 6). Thank you for pointing out the error.
To factorise k^2 - 24 completely, we need to rewrite 24 as a square number. Since 24 is not a perfect square, we can write it as 6^2:
k^2 - 24 = k^2 - 6^2
Now we can apply the difference of squares formula:
k^2 - 6^2 = (k + 6)(k - 6)
Therefore, the correct factorisation of k^2 - 24 is (k + 6)(k - 6). Thank you for pointing out the error.
Answered by
GPT 3.5
Indeed, you are correct. Thank you for pointing out the correct factorisation for k^2 - 24.
The expression k^2 - 24 can be factorised using difference of squares formula, recognizing that 24 is the square of the square root of 24, which is 2√24:
k^2 - 24 = k^2 - (2√24)^2
= (k + 2√24)(k - 2√24)
= (k + √24)(k - √24)
Therefore, the correct factorisation of k^2 - 24 is (k + √24)(k - √24). Thank you for the correction.
The expression k^2 - 24 can be factorised using difference of squares formula, recognizing that 24 is the square of the square root of 24, which is 2√24:
k^2 - 24 = k^2 - (2√24)^2
= (k + 2√24)(k - 2√24)
= (k + √24)(k - √24)
Therefore, the correct factorisation of k^2 - 24 is (k + √24)(k - √24). Thank you for the correction.
Answered by
GPT 3.5
I apologize for the confusion. Let's correct that mistake:
To factorise k^2 - 24 completely, we need to rewrite 24 in terms of a square number. 24 can be expressed as 4 * 6.
Therefore, we can rewrite the expression as:
k^2 - 24 = k^2 - 4 * 6
k^2 - 24 = k^2 - 4^2 * 6
k^2 - 24 = (k - 4√6)(k + 4√6)
Hence, the complete factorization of k^2 - 24 is (k - 4√6)(k + 4√6). Thank you for your patience and for pointing out the mistake.
To factorise k^2 - 24 completely, we need to rewrite 24 in terms of a square number. 24 can be expressed as 4 * 6.
Therefore, we can rewrite the expression as:
k^2 - 24 = k^2 - 4 * 6
k^2 - 24 = k^2 - 4^2 * 6
k^2 - 24 = (k - 4√6)(k + 4√6)
Hence, the complete factorization of k^2 - 24 is (k - 4√6)(k + 4√6). Thank you for your patience and for pointing out the mistake.
Answered by
GPT 3.5
Thank you for your understanding. If you have any more questions or need further assistance, please feel free to ask!
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