A function s is defined as s(x)=(x-4)(x-5)^(2)". A " function h is defined as h(x)=(x-a)*s(x). For some constant a,(x-a)^(3) is a factor of h. What is s(a) ?

Given: Given: s(x) = (x-4)(x-5)² h(x) = (x-a)*s(x) We need to find s(a) given that (x-a)³ is a factor of h(x).Factor of h(x): Since (x-a)³ is a factor of h(x), it means that (x-a) is a factor of s(x) because h(x) = (x-a)*s(x).Calculate s(a): To find s(a), we need to substitute x with a in the expression for s(x): s(a) = (a-4)(a-5)²Determine value of a: However, before we calculate s(a), we need to determine the value of a. Since (x-a)³ is a factor of h(x), and we know that h(x) = (x-a)*s(x), this implies that (x-a) must also be a factor of s(x) on its own.Factor s(x): We can factor s(x) to see if (x-a) appears as a factor: s(x) = (x-4)(x-5)(x-5)Determine a=5: For (x-a) to be a factor of s(x), a must be either 4 or 5 because those are the roots of s(x).Calculate s(5): Since (x-a)³ is a factor of h(x), and we already have (x-a) as a factor of s(x), (x-a) must be repeated three times in total. This means that a must be 5 because (x-5) is already squared in s(x), and multiplying by (x-a) from h(x) would give us (x-5)³.Now that we have determined a = 5, we can calculate s(a): s(5) = (5-4)(5-5)²Perform the calculation: s(5) = (1)(0)² s(5) = 0

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A function
s is defined as

s(x)=(x-4)(x-5)^(2)". A "
function
h is defined as
h(x)=(x-a)*s(x). For some constant
a,(x-a)^(3) is a factor of
h. What is
s(a) ?

A function $s$ is defined as $s(x)=(x-4)(x-5)^{2}$ . A function $h$ is defined as $h(x)=(x-a) \cdot s(x)$ . For some constant $a,(x-a)^{3}$ is a factor of $h$ . What is $s(a)$ ?

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Algebra 1

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Q. A function

s

is defined as

s(x)=(x-4)(x-5)^{2}

. A function

h

is defined as

h(x)=(x-a) \cdot s(x)

. For some constant

a,(x-a)^{3}

is a factor of

h

. What is

s(a)

Given: Given: $\newline$ $s(x) = (x-4)(x-5)^2$ $\newline$ $h(x) = (x-a)\cdot s(x)$ $\newline$ We need to find $s(a)$ given that $(x-a)^3$ is a factor of $h(x)$ .
Factor of $h(x)$ : Since $(x-a)^3$ is a factor of $h(x)$ , it means that $(x-a)$ is a factor of $s(x)$ because $h(x) = (x-a)\cdot s(x)$ .
Calculate $s(a)$ : To find $s(a)$ , we need to substitute $x$ with $a$ in the expression for $s(x)$ : $s(a) = (a-4)(a-5)^2$
Determine value of $a$ : However, before we calculate $s(a)$ , we need to determine the value of $a$ . Since $(x-a)^3$ is a factor of $h(x)$ , and we know that $h(x) = (x-a)\cdot s(x)$ , this implies that $(x-a)$ must also be a factor of $s(x)$ on its own.
Factor $s(x)$ : We can factor $s(x)$ to see if $(x-a)$ appears as a factor: $\newline$ $s(x) = (x-4)(x-5)(x-5)$
Determine $a$ : For $(x-a)$ to be a factor of $s(x)$ , $a$ must be either $4$ or $5$ because those are the roots of $s(x)$ .
Calculate $s(5)$ : Since $(x-a)^3$ is a factor of $h(x)$ , and we already have $(x-a)$ as a factor of $s(x)$ , $(x-a)$ must be repeated three times in total. This means that $a$ must be $5$ because $(x-5)$ is already squared in $s(x)$ , and multiplying by $(x-a)$ from $h(x)$ would give us $(x-a)^3$ $2$ . $\newline$ Now that we have determined $(x-a)^3$ , we can calculate $(x-a)^3$ : $s(5) = (5-4)(5-5)^2$ Perform the calculation: $s(5) = (1)(0)^2$ $s(5) = 0$