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Nonlinear functions SAT question

Math · Advanced Math · Hard · Answer it here, explanation included.

x	p(x)
$x is negative 2$	$p of x is 5$
$x is negative 1$	$p of x is 0$
$x is 0$	$p of x is negative 3$
$x is 1$	$p of x is negative 1$
$x is 2$	$p of x is 0$

The table above gives selected values of a polynomial function p. Based on the values in the table, which of the following must be a factor of p ?

ExplanationShow

Choice D is correct. According to the table, when x is $negative 1$ or 2, $p of x, equals 0$ . Therefore, two x-intercepts of the graph of p are $the points with coordinates negative 1 comma 0$ and $2 comma 0$ . Since $the points with coordinates negative 1 comma 0$ and $2 comma 0$ are x-intercepts, it follows that $open parenthesis, x plus 1, close parenthesis$ and $open parenthesis, x minus 2, close parenthesis$ are factors of the polynomial equation. This is because when $x equals negative 1$ , the value of $x plus 1$ is 0. Similarly, when $x equals 2$ , the value of $x minus 2$ is 0. Therefore, the product $open parenthesis, x plus 1, close parenthesis, times, open parenthesis, x minus 2, close parenthesis$ is a factor of the polynomial function p.

Choice A is incorrect. The factor $x minus 3$ corresponds to an x-intercept of $the point with coordinates 3 comma 0$ , which isn’t present in the table. Choice B is incorrect. The factor $x plus 3$ corresponds to an x-intercept of $the point with coordinates negative 3 comma 0$ , which isn’t present in the table. Choice C is incorrect. The factors $x minus 1$ and $x plus 2$ correspond to x-intercepts $with coordinates 1 comma 0$ and $negative 2 comma 0$ , respectively, which aren’t present in the table.

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