factorise 2a3+10a2-28a
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Answered by
109
2a³ + 10a² - 28a
2a(a²+5a-28)
2a[(a²+7a-2a-14)]
2a[ a(a+7)-2(a+7)]
2a[ (a+7) (a-2) ]
2a(a²+5a-28)
2a[(a²+7a-2a-14)]
2a[ a(a+7)-2(a+7)]
2a[ (a+7) (a-2) ]
hercules4:
its 28 not 8×3=24
Answered by
55
2a^3+10a^2-28a
take 2a as common
(2a)(a^2+5a-14)
middle term split of second factor
a^2+7a-2a-14
a(a+7)-2(a+7)
(a-2)(a+7)
so three factors are 2a,a-2,a+7
take 2a as common
(2a)(a^2+5a-14)
middle term split of second factor
a^2+7a-2a-14
a(a+7)-2(a+7)
(a-2)(a+7)
so three factors are 2a,a-2,a+7
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