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MakcuM [25]
3 years ago
12

given the trinomial x^2+bx-c where both the first sign is positive and the second sign is negative, the signs of the factors wil

l be:
Mathematics
2 answers:
KatRina [158]3 years ago
7 0

Answer one positive one negative

geniusboy [140]3 years ago
5 0

The signs of the factors will be different. (One will be positive and one will be negative.)

_____

Here, the "first sign" is considered to be the sign of <em>x²</em>, and the "second sign" is considered to be the sign of <em>c</em>. The above statement will be true regardless of the sign of <em>b</em>. The "signs <em>of</em> the factors" is considered to refer to the signs of the constant terms <em>in</em> the binomial factors.

For (x +p)(x -q) where <em>p</em> and <em>q</em> are both positive so the signs are as shown, the product is x² +(p-q)x -pq. That is <em>x²</em> is positive and <em>-pq</em> is negative, regardless of the relative magnitudes of <em>p</em> and <em>q</em> (thus the sign of <em>p-q</em>).

_____

If something else is meant by the terminology used, it isn't clear what the intended answer is supposed to be.

Both x² +3x -10 and x² -3x -10 have factorizations that have different signs <em>in</em> the two factors. The first is (x+5)(x-2); the second is (x+2)(x-5). The signs <em>of</em> the factors are all positive: (x+5), not -(x+5), for example.

On the other hand x² -3x +2 = (x-2)(x-1) has same signs <em>in</em> and <em>of</em> the factors. That is, by themselves, the sign of x² (first sign) and the sign of 3x (second sign) don't guarantee the signs <em>in</em> the factors are anything in particular, except that at least one sign in the factors must be negative if the first (x²) and second (3x) signs differ.

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