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secondary 4 | A Maths
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Hi! I am doing a question on differentiation and I’m confused as to how to arrive at the last step from the second last step. Thank you so much if you helped :)
Date Posted: 4 years ago
Views: 346
We simply take out the common terms just like we normally do in standard common term factorisation.
do u mean this?
(x - 5)⁷ [4(x + 3)³] + (x + 3)⁴ [7(x - 5)³]
= (x - 5)⁶ (x + 3)³ [ 4(x - 5) ]
+ (x + 3)³ (x - 5)⁶ [7(x + 3)]
= (x - 5)⁶ (x + 3)³ (4x - 20)
+ (x + 3)³ (x - 5)⁶ (7x + 21)
= (x - 5)⁶ (x + 3)³ (4x - 20 + 7x + 21)
= (x - 5)⁶ (x + 3)³ (11x + 1)
.
.
.
(x - 5)⁷ [4(x + 3)³] + (x + 3)⁴ [7(x - 5)³]
= (x - 5)⁶ (x + 3)³ [ 4(x - 5) ]
+ (x + 3)³ (x - 5)⁶ [7(x + 3)]
= (x - 5)⁶ (x + 3)³ (4x - 20)
+ (x + 3)³ (x - 5)⁶ (7x + 21)
= (x - 5)⁶ (x + 3)³ (4x - 20 + 7x + 21)
= (x - 5)⁶ (x + 3)³ (11x + 1)
.
.
.
thank you guys
See 2 Answers
You missed the point fully. We are supposed to factorise the final expression further to the simplest form.
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done
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It’s basically your standard factorisation techniques, except that the expressions are more complicated this time.
Date Posted:
4 years ago
Thank you!
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