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Which point is a point where the graph of [tex]y = (x-5)\left(x^2 - 7x + 12\right)[/tex] crosses the [tex]x[/tex]-axis?

A. [tex](-5,0)[/tex]
B. [tex](-3,0)[/tex]
C. [tex](4,0)[/tex]
D. [tex](12,0)[/tex]

Sagot :

To determine the points where the graph of the polynomial [tex]\( y = (x-5)(x^2 - 7x + 12) \)[/tex] crosses the [tex]\( x \)[/tex]-axis, we need to find the roots of the polynomial, i.e., the values of [tex]\( x \)[/tex] that make [tex]\( y \)[/tex] equal to zero.

1. Step 1: Factor the polynomial completely.

First, we expand the polynomial:
[tex]\[ y = (x-5)(x^2 - 7x + 12) \][/tex]

Next, factor the quadratic part [tex]\( x^2 - 7x + 12 \)[/tex]. To factor [tex]\( x^2 - 7x + 12 \)[/tex], we look for two numbers that multiply to 12 and add to -7. These numbers are -3 and -4. Therefore, we can factor the quadratic expression as:
[tex]\[ x^2 - 7x + 12 = (x-3)(x-4) \][/tex]

2. Step 2: Rewrite the entire polynomial with the factors.

Substituting the factored form of the quadratic part back into the polynomial, we get:
[tex]\[ y = (x-5)(x-3)(x-4) \][/tex]

3. Step 3: Identify the roots of the polynomial.

The polynomial [tex]\( y = (x-5)(x-3)(x-4) \)[/tex] is zero when any of the factors are zero. Thus, we find the roots by setting each factor equal to zero:
- [tex]\( x - 5 = 0 \)[/tex] gives [tex]\( x = 5 \)[/tex]
- [tex]\( x - 3 = 0 \)[/tex] gives [tex]\( x = 3 \)[/tex]
- [tex]\( x - 4 = 0 \)[/tex] gives [tex]\( x = 4 \)[/tex]

These are the [tex]\( x \)[/tex]-values where the polynomial equals zero. The points where the graph crosses the [tex]\( x \)[/tex]-axis are therefore:
- [tex]\( (5, 0) \)[/tex]
- [tex]\( (3, 0) \)[/tex]
- [tex]\( (4, 0) \)[/tex]

4. Step 4: Select the correct point from the given options.

From the options provided:
- [tex]\( (-5, 0) \)[/tex]
- [tex]\( (-3, 0) \)[/tex]
- [tex]\( (4, 0) \)[/tex]
- [tex]\( (12, 0) \)[/tex]

We see that the correct point among the choices is [tex]\( (4, 0) \)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{(4,0)} \][/tex]