Answer :
To factorize the trinomial [tex]\(x^2 - 7x + 6\)[/tex], we need to express it as a product of two binomials of the form [tex]\((x + a)(x + b)\)[/tex].
The first term in the trinomial, [tex]\(x^2\)[/tex], indicates that the first terms in each binomial must be [tex]\(x\)[/tex]. So, our binomials will look like this:
[tex]\[ (x + a)(x + b) \][/tex]
Expanding [tex]\((x + a)(x + b)\)[/tex] gives:
[tex]\[ x^2 + (a + b)x + ab \][/tex]
We need the expanded form to match [tex]\(x^2 - 7x + 6\)[/tex]. Therefore, we must find constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
[tex]\[ a + b = -7 \quad \text{and} \quad ab = 6 \][/tex]
Next, we look for pairs of integers that multiply to 6:
- [tex]\(1 \cdot 6 = 6\)[/tex]
- [tex]\((-1) \cdot (-6) = 6\)[/tex]
- [tex]\(2 \cdot 3 = 6\)[/tex]
- [tex]\((-2) \cdot (-3) = 6\)[/tex]
Among these pairs, the pair [tex]\((-6)\)[/tex] and [tex]\((-1)\)[/tex] sums up to [tex]\(-7\)[/tex]:
[tex]\[ -6 + (-1) = -7 \][/tex]
So, the correct factorization is:
[tex]\[ (x - 6)(x - 1) \][/tex]
Hence, the answer is:
[tex]\[ D. \ (x - 6)(x - 1) \][/tex]
The first term in the trinomial, [tex]\(x^2\)[/tex], indicates that the first terms in each binomial must be [tex]\(x\)[/tex]. So, our binomials will look like this:
[tex]\[ (x + a)(x + b) \][/tex]
Expanding [tex]\((x + a)(x + b)\)[/tex] gives:
[tex]\[ x^2 + (a + b)x + ab \][/tex]
We need the expanded form to match [tex]\(x^2 - 7x + 6\)[/tex]. Therefore, we must find constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that:
[tex]\[ a + b = -7 \quad \text{and} \quad ab = 6 \][/tex]
Next, we look for pairs of integers that multiply to 6:
- [tex]\(1 \cdot 6 = 6\)[/tex]
- [tex]\((-1) \cdot (-6) = 6\)[/tex]
- [tex]\(2 \cdot 3 = 6\)[/tex]
- [tex]\((-2) \cdot (-3) = 6\)[/tex]
Among these pairs, the pair [tex]\((-6)\)[/tex] and [tex]\((-1)\)[/tex] sums up to [tex]\(-7\)[/tex]:
[tex]\[ -6 + (-1) = -7 \][/tex]
So, the correct factorization is:
[tex]\[ (x - 6)(x - 1) \][/tex]
Hence, the answer is:
[tex]\[ D. \ (x - 6)(x - 1) \][/tex]
