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Factor the trinomial below.

[tex]\[ x^2 - 14x + 45 \][/tex]

A. [tex][tex]$(x-5)(x-9)$[/tex][/tex]

B. [tex][tex]$(x-5)(x+9)$[/tex][/tex]

C. [tex][tex]$(x-3)(x+15)$[/tex][/tex]

D. [tex][tex]$(x-3)(x-15)$[/tex][/tex]

Sagot :

To factor the given trinomial [tex]\(x^2 - 14x + 45\)[/tex], let's follow a systematic approach:

1. Identify the coefficients: For the quadratic trinomial [tex]\(ax^2 + bx + c\)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -14 \)[/tex]
- [tex]\( c = 45 \)[/tex]

2. Find two numbers that multiply to [tex]\(ac\)[/tex] and add to [tex]\(b\)[/tex]:
- We need two numbers that multiply to [tex]\(a \cdot c = 1 \cdot 45 = 45\)[/tex] and add to [tex]\(b = -14\)[/tex].

3. List pairs of factors of 45:
- [tex]\(1 \times 45\)[/tex]
- [tex]\(3 \times 15\)[/tex]
- [tex]\(5 \times 9\)[/tex]

4. Determine which pair sums to -14:
- Check if [tex]\(1 + 45 = 46\)[/tex]
- Check if [tex]\(3 + 15 = 18\)[/tex]
- Check if [tex]\(5 + 9 = 14\)[/tex], and we find that [tex]\(9 + 5 = 14\)[/tex]. Now, since [tex]\(b\)[/tex] is negative, we will consider [tex]\(-9\)[/tex] and [tex]\(-5\)[/tex].

5. Write the factors:
- The numbers that match our requirements are [tex]\(-9\)[/tex] and [tex]\(-5\)[/tex].

6. Factor the trinomial:
- Thus, [tex]\(x^2 - 14x + 45\)[/tex] can be written as [tex]\((x - 9)(x - 5)\)[/tex].

Therefore, the trinomial [tex]\(x^2 - 14x + 45\)[/tex] factors to:
[tex]\[ (x - 9)(x - 5) \][/tex]

Let's identify the correct multiple-choice answer:
A. [tex]\((x - 5)(x - 9)\)[/tex]
B. [tex]\((x - 5)(x + 9)\)[/tex]
C. [tex]\((x - 3)(x + 15)\)[/tex]
D. [tex]\((x - 3)(x - 15)\)[/tex]

Given that multiplication is commutative, [tex]\( (x - 9)(x - 5) \)[/tex] is the same as [tex]\( (x - 5)(x - 9) \)[/tex].

Thus, the correct answer is:
[tex]\[ \boxed{A} \][/tex]