Find all real zeros of the function
x^3 +14x^2+41x-56

Question

Find all real zeros of the function
x^3 +14x^2+41x-56

anonymous · Answer

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anonymous · Answer

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anonymous · Answer

If $a,b,c$ are zeros of the polynomial, then you should be able to write it as the product
$$(x-a)(x-b)(x-c)=x^3+14x^2+41x-56$$
Notice that the product $(-a)(-b)(-c)=-56$, so $a,b,c$ form a triplet of factors of -56. There may be a more efficient way of doing this, but since 56 isn't such a large number, we can list the factors: $1,2,4,7,8,14,28,56$.

We can sort out the plausible triplets: $(1,1,56)$, $(1,2,28)$, $(1,4,14)$, $(1,7,8)$, $(2,2,14)$, and $(2,4,7)$. (This was done using the prime factorization of 56, which is $2^37^1$, which is then partitioned into 3 factors.) Of these, you will also have check a variety of positive/negative factors. Very tedious.

I've checked the answer, and since the roots are integers there are a few ways you can rearrange the given polynomial's terms to be able to implement a factorization by grouping.