1
Question
Factorise :
3−a(4+7a)
Factorise :
3−a(4+7a)
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Solution
3−a(4+7a)=3−4a−7a2=−1(7a2+4a−3)=−1(7a2+7a−3a−3)=−1[7a(a+1)−3(a+1)]=−1(a+1)(7a−3)=(a+1)(3−7a)
3−a(4+7a)=3−4a−7a2=−1(7a2+4a−3)=−1(7a2+7a−3a−3)=−1[7a(a+1)−3(a+1)]=−1(a+1)(7a−3)=(a+1)(3−7a)
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