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).
Factorise k^2-24 completely
11 answers
but in my world
(k-6)(k+4) = k^2 -2 k - 24 ,
not k^2 - 24
(k-6)(k+4) = k^2 -2 k - 24 ,
not k^2 - 24
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).
Sorry, 6*6 = 36 not 24
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.
k^2 - 24 = (k+√24)(k-√24)
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.
You still have the 6*6 = 36 not 24 problem
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.
Yes :)
Thank you for your understanding. If you have any more questions or need further assistance, please feel free to ask!