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secondary 3 | A Maths
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Pls help with q12 thanks
Date Posted: 4 years ago
Views: 475
Numerator's degree of polynomial is higher than denominator. So do division first.
(2x³ - 2x² - 4x + 2) / (2x² - 8)
= (x³ - x² - 2x + 1) / (x² - 4)
= ( (x² - 4)(x - 1) - 3 + 2x ) /(x² - 4)
(You can do this step mentally or long division by (x² - 4) )
= x - 1 + (2x - 3)/(x² - 4)
= x - 1 + (2x - 3) / ( (x + 2)(x - 2) )
Now,
Let (2x - 3) / ( (x + 2)(x - 2) ) = A/(x + 2) + B/(x - 2)
By the cover up rule,
A = (2(-2) - 3) / (-2 - 2) = -7/-4 = 7/4
B = (2(2) - 3) / (2 + 2) = 1/4
So,
(2x³ - 2x² - 4x + 2) / (2x² - 8)
= x - 1 + 7/(4(x + 2)) + 1/(4(x - 2))
(2x³ - 2x² - 4x + 2) / (2x² - 8)
= (x³ - x² - 2x + 1) / (x² - 4)
= ( (x² - 4)(x - 1) - 3 + 2x ) /(x² - 4)
(You can do this step mentally or long division by (x² - 4) )
= x - 1 + (2x - 3)/(x² - 4)
= x - 1 + (2x - 3) / ( (x + 2)(x - 2) )
Now,
Let (2x - 3) / ( (x + 2)(x - 2) ) = A/(x + 2) + B/(x - 2)
By the cover up rule,
A = (2(-2) - 3) / (-2 - 2) = -7/-4 = 7/4
B = (2(2) - 3) / (2 + 2) = 1/4
So,
(2x³ - 2x² - 4x + 2) / (2x² - 8)
= x - 1 + 7/(4(x + 2)) + 1/(4(x - 2))
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Date Posted:
4 years ago
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