Matthew Fan's answer to hfrh's Secondary 3 A Maths Singapore question.
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x^3+mx^2+nx-6=(x^2-x-6)(ax+b)
(i) Note that x^2-x-6=(x-3)(x+2). Hence when x=3 and x=-2, LHS is 0. So we put the values into LHS to get 27+9m+3n-6=0, which simplifies into 3m+n=-7 and -8+4m-2n-6=0, which simplifies into 2m-n=7
Solve simultaneous equations to get m=0, n=-7
(ii) In the equation, the RHS expanded out has coefficient of x^3 as 'a' and coefficient of constant as '-6b'. Comparing with LHS gives a=1 and b=1. Hence the full factorisation is x^3-7x-6=(x+2)(x-3)(x+1)
(i) Note that x^2-x-6=(x-3)(x+2). Hence when x=3 and x=-2, LHS is 0. So we put the values into LHS to get 27+9m+3n-6=0, which simplifies into 3m+n=-7 and -8+4m-2n-6=0, which simplifies into 2m-n=7
Solve simultaneous equations to get m=0, n=-7
(ii) In the equation, the RHS expanded out has coefficient of x^3 as 'a' and coefficient of constant as '-6b'. Comparing with LHS gives a=1 and b=1. Hence the full factorisation is x^3-7x-6=(x+2)(x-3)(x+1)
Date Posted:
4 years ago
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