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Sagot :
To determine which trinomial was factored to match the factorized form provided, let's examine each of the trinomials given in the problem one by one and compare them to the factorized forms presented in the results: [tex]\((x - 2)(x + 6)\)[/tex], [tex]\(x^2 + 2x - 6\)[/tex], [tex]\(x^2 - 8x - 6\)[/tex], and [tex]\((x - 6)(x + 2)\)[/tex].
1. For the trinomial [tex]\(x^2 + 4x - 12\)[/tex]:
To factorize [tex]\(x^2 + 4x - 12\)[/tex], we look for two numbers that multiply to [tex]\(-12\)[/tex] (the constant term) and add to [tex]\(4\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term). Those numbers are [tex]\(6\)[/tex] and [tex]\(-2\)[/tex]. Therefore, the factorized form of [tex]\(x^2 + 4x - 12\)[/tex] is:
[tex]\[(x + 6)(x - 2)\][/tex]
2. For the trinomial [tex]\(x^2 + 2x - 6\)[/tex]:
To factorize [tex]\(x^2 + 2x - 6\)[/tex], we look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(2\)[/tex]. There are no such pairs of integers. Therefore, the trinomial [tex]\(x^2 + 2x - 6\)[/tex] does not factorize nicely into integer factors and stays as it is:
[tex]\[x^2 + 2x - 6\][/tex]
3. For the trinomial [tex]\(x^2 - 8x - 6\)[/tex]:
To factorize [tex]\(x^2 - 8x - 6\)[/tex], we look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(-8\)[/tex]. There are no such pairs of integers. Therefore, the trinomial [tex]\(x^2 - 8x - 6\)[/tex] does not factorize nicely into integer factors and stays as it is:
[tex]\[x^2 - 8x - 6\][/tex]
4. For the trinomial [tex]\(x^2 - 4x - 12\)[/tex]:
To factorize [tex]\(x^2 - 4x - 12\)[/tex], we look for two numbers that multiply to [tex]\(-12\)[/tex] and add to [tex]\(-4\)[/tex]. Those numbers are [tex]\(2\)[/tex] and [tex]\(-6\)[/tex]. Therefore, the factorized form of [tex]\(x^2 - 4x - 12\)[/tex] is:
[tex]\[(x + 2)(x - 6)\][/tex]
Given these factorizations:
- [tex]\(x^2 + 4x - 12\)[/tex] factorizes to [tex]\((x - 2)(x + 6)\)[/tex]
- [tex]\(x^2 + 2x - 6\)[/tex] does not factorize nicely into integer factors, remains [tex]\(x^2 + 2x - 6\)[/tex]
- [tex]\(x^2 - 8x - 6\)[/tex] does not factorize nicely into integer factors, remains [tex]\(x^2 - 8x - 6\)[/tex]
- [tex]\(x^2 - 4x - 12\)[/tex] factorizes to [tex]\((x - 6)(x + 2)\)[/tex]
Thus, the trinomial [tex]\(x^2 + 4x - 12\)[/tex] was factored to [tex]\((x - 2)(x + 6)\)[/tex]. Therefore, the trinomial that was factored is:
[tex]\[ x^2 + 4x - 12 \][/tex]
1. For the trinomial [tex]\(x^2 + 4x - 12\)[/tex]:
To factorize [tex]\(x^2 + 4x - 12\)[/tex], we look for two numbers that multiply to [tex]\(-12\)[/tex] (the constant term) and add to [tex]\(4\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term). Those numbers are [tex]\(6\)[/tex] and [tex]\(-2\)[/tex]. Therefore, the factorized form of [tex]\(x^2 + 4x - 12\)[/tex] is:
[tex]\[(x + 6)(x - 2)\][/tex]
2. For the trinomial [tex]\(x^2 + 2x - 6\)[/tex]:
To factorize [tex]\(x^2 + 2x - 6\)[/tex], we look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(2\)[/tex]. There are no such pairs of integers. Therefore, the trinomial [tex]\(x^2 + 2x - 6\)[/tex] does not factorize nicely into integer factors and stays as it is:
[tex]\[x^2 + 2x - 6\][/tex]
3. For the trinomial [tex]\(x^2 - 8x - 6\)[/tex]:
To factorize [tex]\(x^2 - 8x - 6\)[/tex], we look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(-8\)[/tex]. There are no such pairs of integers. Therefore, the trinomial [tex]\(x^2 - 8x - 6\)[/tex] does not factorize nicely into integer factors and stays as it is:
[tex]\[x^2 - 8x - 6\][/tex]
4. For the trinomial [tex]\(x^2 - 4x - 12\)[/tex]:
To factorize [tex]\(x^2 - 4x - 12\)[/tex], we look for two numbers that multiply to [tex]\(-12\)[/tex] and add to [tex]\(-4\)[/tex]. Those numbers are [tex]\(2\)[/tex] and [tex]\(-6\)[/tex]. Therefore, the factorized form of [tex]\(x^2 - 4x - 12\)[/tex] is:
[tex]\[(x + 2)(x - 6)\][/tex]
Given these factorizations:
- [tex]\(x^2 + 4x - 12\)[/tex] factorizes to [tex]\((x - 2)(x + 6)\)[/tex]
- [tex]\(x^2 + 2x - 6\)[/tex] does not factorize nicely into integer factors, remains [tex]\(x^2 + 2x - 6\)[/tex]
- [tex]\(x^2 - 8x - 6\)[/tex] does not factorize nicely into integer factors, remains [tex]\(x^2 - 8x - 6\)[/tex]
- [tex]\(x^2 - 4x - 12\)[/tex] factorizes to [tex]\((x - 6)(x + 2)\)[/tex]
Thus, the trinomial [tex]\(x^2 + 4x - 12\)[/tex] was factored to [tex]\((x - 2)(x + 6)\)[/tex]. Therefore, the trinomial that was factored is:
[tex]\[ x^2 + 4x - 12 \][/tex]
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