When $f(x)$ is divided by $x-1$ , the remainder is $12$ . When $f(x)$ is divided by $x-4$ , the remainder is $3$ . Find the remainder when $f(x)$ is divided by $x^{2}-5 x+4$ .

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Algebra 1

Unions and intersections of sets

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Q. When

f(x)

is divided by

x-1

, the remainder is

12

. When

f(x)

is divided by

x-4

, the remainder is

3

. Find the remainder when

f(x)

is divided by

x^{2}-5 x+4

Understand the Problem: Understand the problem. $\newline$ We are given that $f(x)$ leaves a remainder of $12$ when divided by $x - 1$ , and a remainder of $3$ when divided by $x - 4$ . We need to find the remainder when $f(x)$ is divided by $(x - 1)(x - 4) = x^2 - 5x + 4$ .
Use the Remainder Theorem: Use the Remainder Theorem. $\newline$ The Remainder Theorem states that if a polynomial $f(x)$ is divided by $x - a$ , the remainder is $f(a)$ . We apply this theorem to the given information. $\newline$ For $x - 1$ : $f(1) = 12$ $\newline$ For $x - 4$ : $f(4) = 3$
Express $f(x)$ : Express $f(x)$ in terms of the divisors and remainders. $\newline$ Since we know the remainders when $f(x)$ is divided by $x - 1$ and $x - 4$ , we can express $f(x)$ as: $\newline$ $f(x) = (x - 1)Q_1(x) + 12 = (x - 4)Q_2(x) + 3$ $\newline$ where $Q_1(x)$ and $Q_2(x)$ are some quotient polynomials.
Find Relationship: Find the relationship between $Q1(x)$ and $Q2(x)$ . $\newline$ Since both $(x - 1)Q1(x) + 12$ and $(x - 4)Q2(x) + 3$ represent the same polynomial $f(x)$ , and the remainders are given for specific values of $x$ , we can equate the two expressions for those values of $x$ . $\newline$ For $x = 1$ : $(1 - 1)Q1(1) + 12 = 12$ $\newline$ For $x = 4$ : $Q2(x)$ $0$ $\newline$ This does not give us information about $Q1(x)$ and $Q2(x)$ , but it confirms the remainders are correct.
Find Remainder: Find the remainder when $f(x)$ is divided by $(x - 1)(x - 4)$ . We need to find a polynomial $R(x)$ of degree less than $2$ (since the divisor is a quadratic) such that: $f(x) = (x^2 - 5x + 4)Q(x) + R(x)$ where $Q(x)$ is the quotient polynomial and $R(x)$ is the remainder polynomial we are looking for.
Use Given Remainders: Use the given remainders to determine $R(x)$ . $\newline$ Since $R(x)$ must give the same remainders as $f(x)$ for $x = 1$ and $x = 4$ , we can set up a system of equations: $\newline$ $R(1) = 12$ $\newline$ $R(4) = 3$ $\newline$ Assuming $R(x)$ is of the form $ax + b$ , we can write: $\newline$ $a(1) + b = 12$ $\newline$ $R(x)$ $0$
Solve Equations: Solve the system of equations for $a$ and $b$ . From the first equation, we get: $a + b = 12$ From the second equation, we get: $4a + b = 3$ Subtracting the first equation from the second, we get: $3a = -9$ $a = -3$ Substituting $a$ back into the first equation: $-3 + b = 12$ $b = 15$
Write Remainder Polynomial: Write down the remainder polynomial $R(x)$ . We found that $a = -3$ and $b = 15$ , so the remainder polynomial $R(x)$ is: $R(x) = -3x + 15$