2023 Mathematics Competition

### Question 22

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### 6 comments

###### Akinola Ajayi

5 months agoConsidering 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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