2023 Mathematics Competition
Question 22
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6 comments
Akinola Ajayi
1 year ago
Considering the cubic equation above
x³+3x²+4x-12=0
Let the values be l, m, n
x³+3x²+4x-12=(x-l)(x-m)(x-n)
By expanding the RHS, we have:
(x-l)(x-m)(x-n) = x³-(l+m+n)x² + (lm+ln+mn)x - lmn
Thus,
x³+3x²+4x-12=x³-(l+m+n)x² + (lm+ln+mn)x - lmn
By comparing the coefficient, we have:
Sum of the values for x to be
l+m+n=-3
Sum of the product of the values for x to be;
lm+ln+mn=4
Product of the values of x is given by
-lmn=-12
lmn=12
Therefore the product of the values for x in the equation x³+3x²+4x-12 is 12
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