The table below shows selected values of x and the corresponding values of the polynomial function p(x). | x | p(x) | | :--- | :--- | | 0 | -16 | | 2 | 0 | | 3 | -4 | | 4 | 0 | If p(x) = a(x-2)(x-4)(x-k) for all real numbers x, where a and k are constants, what is the value of k?

The table below shows selected values of x and the corresponding values of the polynomial … — SAT

The table below shows selected values of $x$ and the corresponding values of the polynomial function $p(x)$ .

$x$	$p(x)$
$0$	$-16$
$2$	$0$
$3$	$-4$
$4$	$0$

If $p(x) = a(x-2)(x-4)(x-k)$ for all real numbers $x$ , where $a$ and $k$ are constants, what is the value of $k$ ?

Answer: 1

Answer

The value of

k

1

Substituting the coordinate pair

(0, -16)

into the equation

p(x) = a(x-2)(x-4)(x-k)

gives

p(0) = a(-2)(-4)(-k) = -8ak = -16

, which simplifies to

ak = 2

. Substituting the coordinate pair

(3, -4)

gives

p(3) = a(3-2)(3-4)(3-k) = -a(3-k) = -4

. Expanding this equation yields

-3a + ak = -4

. Replacing

ak

with

2

in this equation results in

-3a + 2 = -4

, which simplifies to

-3a = -6

, so

a = 2

. Using

a = 2

in the relation

ak = 2

gives

2k = 2

, which yields

k = 1

Step-by-Step Solution

Substitute the coordinates

(0, -16)

from the table into the given equation

p(x) = a(x-2)(x-4)(x-k)

to form a relationship between the constants.

p(0) = a(0-2)(0-4)(0-k) = a(-2)(-4)(-k) = -8ak = -16

, which simplifies to

ak = 2

To establish a relation between constants

a

and

k

using the y-intercept of the polynomial.

Substitute another known point from the table,

(3, -4)

, into the polynomial equation.

p(3) = a(3-2)(3-4)(3-k) = a(1)(-1)(3-k) = -a(3-k) = -4

To obtain a second equation relating the constants

a

and

k

Solve the system of equations by distributing

-a

in the second equation and substituting the value of

ak

from the first step.

-3a + ak = -4

. Since

ak = 2

, this becomes

-3a + 2 = -4 \implies -3a = -6 \implies a = 2

To solve for the leading coefficient constant

a

Substitute the value of

a

back into the first relation to solve for

k

Since

a = 2

and

ak = 2

, we have

2k = 2 \implies k = 1

To find the final value of

k

Key Concept

Solving for polynomial constants using given coordinate points from a table

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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