The polynomial has a factor and when divided by , a
remainder of -54 is obtained. Find the values of and . With these values of and ,
factorise the polynomial completely. Hence, or otherwise,
solve .
[Answer Key]
Use the Factor Theorem: Since is a factor of the polynomial , we can substitute into the polynomial and set it equal to zero.
Use the Remainder Theorem: Since the remainder when divided by is -54, we can substitute into the polynomial and set it equal to -54.
Set up the equations: From the above two steps, we will derive two equations involving and .
Solve the system of equations: We will solve the equations to find the values of and .
Factor the polynomial: With the values of and , we will factor the polynomial completely.
Solve the equation using the factors obtained.
Let’s start with the first two steps:
Step 1: Substitute
This simplifies to:
Step 2: Substitute
This simplifies to:
Step 3: Set up the equations
Now we have two equations:
Step 4: Solve the system of equations
Setting the two expressions for equal to each other:
Rearranging gives:
Now substituting back into Equation 1 to find :
Thus, we have:
Step 5: Factor the polynomial
Now substituting and back into the polynomial:
We know is a factor, so we can perform polynomial long division or synthetic division to factor it completely.
Let’s perform synthetic division with :
The result is .
Now we need to factor . We can use the quadratic formula:
This gives us:
Thus, the complete factorization of the polynomial is:
Step 6: Solve
We can substitute :
Using the factors we found, we can substitute back to find the roots:
gives or
gives or
Thus, the solutions to are:
Final Answers
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Given that is a factor, we can apply the factor theorem, which states that if is a factor of a polynomial , then . This means we need to substitute into the polynomial:
Setting :
Next, considering the condition that gives a remainder of -54, we substitute :
Setting this equal to -54:
Now, we have a system of equations from (1) and (2):
Next, we solve for :
From (1):
Substituting into (2):
Now, substituting back to find :
Thus, we have and .
Now we can express the polynomial with these values:
To factor , we have already established it has a factor . To find the complete factorization, we can perform polynomial long division or synthetic division of by :