5. Dividing Polynomials
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Chapter 1
5.
Dividing Polynomials
| Student Learning Objectives: |
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Why
Polynomials, expressions involving terms raised to a power, are fundamental in mathematics. Dividing one polynomial by another is a common task that requires specific techniques, one of which is polynomial long division. This method mirrors the traditional long division we use with numbers but applies it to polynomial expressions. Understanding this technique is crucial for solving complex mathematical problems, especially in algebra. It aids in simplifying functions and evaluating certain limits, making tasks more manageable and solutions clearer.
Lesson Settings & Tools
| | 11 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Dividing Polynomials
Slide of 11
Two polynomials can be added or subtracted by combining their like terms. Additionally, the polynomials can be multiplied by using either the FOIL method, the Box Method, or the Distributive Property. Notice that the result of any of these operations is always a polynomial. Therefore, the set of polynomials is closed under addition, subtraction, and multiplication.
| Operation | Methods | Result |
|---|---|---|
| Addition | Combining Like Terms | Polynomial |
| Subtraction | Combining Like Terms | Polynomial |
| Multiplication | The FOIL Method The Box Method The Distributive Property |
Polynomial |
This lesson completes the set of the four basic operations for polynomials by investigating the methods to divide two polynomials.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Analyzing the Product of a Polynomial and (x-k)
Consider the polynomials P(x) and Q(x) and the constant k.
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Discussion
Dividing Polynomials
Adding, subtracting, and multiplying polynomials are operations that can be performed by applying the properties of real numbers and the product of powers property and then combining like terms. However, dividing polynomials may not be so intuitive. 3x^3+2x^2-7/x^2+2x+5 = ? The good news is that polynomial division can be performed similarly to dividing integer numbers. In long division notation, the numerator is placed underneath the division symbol and the denominator is placed to the left. As a refresher, the process for calculating 117 4 is shown.
Method
Polynomial Long Division
To perform the division of two polynomials, the degree of the numerator must be greater than or equal to the degree of the denominator. For example, consider the following division. 3x^3+2x^2-7/x^2+2x+5 The process for finding the division of two polynomials is similar to the process of dividing integers and is summarized in the following five steps.
1
Sort the Terms and Write as Long Division
expand_more To begin, write both polynomials in standard form. If the numerator has missing terms, include them with coefficient 0. For example, the numerator will be written as 3x^3+2x^2+ 0x-7. Then, write the polynomials as in long division notation.
x^2+2x+5 & |r 3x^3+2x^2+ 0x-7
2
Divide the Leading Terms
expand_more Divide the leading term of the numerator by the leading term of the denominator. In this case, divide 3x^3 by x^2.
3x^3/x^2 = 3x
Write the resulting monomial above the horizontal bar.
rr & 3x x^2+2x+5 & |r 3x^3+2x^2+0x-7
3
Multiply the Result From Step 2 by the Denominator
expand_more The result from step 2 is 3x. Therefore, multiply 3x by the denominator x^2+2x+5.
3x( x^2+2x+5) = 3x^3 + 6x^2 + 15x
Next, write the resulting expression below the numerator and try to align the common powers.
rr & 3x x^2+2x+5 & |r 3x^3+2x^2+0x-7 & c 3x^3+6x^2+15x
4
Subtract the Product From Step 3 From the Numerator
expand_more Subtract the polynomial obtained in step 3 from the numerator.
rr & 3x x^2+2x+5 & |r 3x^3+2x^2+0x-7 & c -( 3x^3+6x^2+15x) & c -4x^2-15x-7
The result of the subtraction — the remainder — is the polynomial -4x^2-15x-7. This means that the division of the given polynomials can be written as follows.
3x^3+2x^2-7/x^2+2x+5 = 3x + -4x^2-15x-7/x^2+2x+5
However, since the degree of -4x^2-15x-7 is the same as the degree of the denominator, the division is not done yet. Thus, repeat steps 2-4 until the remainder's degree is less than the denominator's degree.
5
Repeat Steps 2-4 Until the Remainder's Degree is Less Than the Denominator's Degree
expand_more Steps 2-4 are repeated until the polynomial obtained in step 4 has a lower degree than the denominator. Dividing the leading terms -4x^2 and x^2 gives -4 as result. Next, multiply the denominator by -4 and write the result at the bottom.
rr & 3x-4 x^2+2x+5 & |r 3x^3+2x^2+0x-7 & c -(3x^3+6x^2+15x) & c -4x^2-15x-7 & c -4x^2-8x-20
Now, subtract the two polynomials at the bottom.
rr & 3x-4 x^2+2x+5 & |r 3x^3+2x^2+0x-7 & c -(3x^3+6x^2+15x) & c -4x^2-15x-7 & c -( -4x^2-8x-20) & c -7x+13
This time, the resulting polynomial has a degree of 1, which is lower than the denominator's degree. Therefore, the division is complete. This implies that the quotient of the division is 3x-4 and the remainder is -7x+13.
Q(x) &= 3x-4 R(x) &= -7x+13
Finally, the initial polynomial division can be written as the quotient plus the division of the remainder and the denominator.
3x^3+2x^2-7/x^2+2x+5 = 3x-4 + -7x+13/x^2+2x+5
Extra
What Happens When the Numerator's Degree Is Less Than the Denominator's Degree?When the numerator has a lower degree than the denominator, the result from dividing the leading terms will have a negative exponent. Therefore, it will not be a term of a polynomial. For example, consider the following division.
4x+1/x^2-3
The result of dividing the leading terms is 4x^(-1), which is not a monomial.
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Example
Voltage of a Circuit
Electrical power is defined as the product of the voltage of a circuit and the current flowing through it. In other words, electrical power equals voltage times current.
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Hint
Divide the polynomials using long division.
Solution
According to the formula, the voltage of a circuit is found by dividing the power by the current.
Voltage = Power/Current
In this case, both the power and the current of the circuit are represented by polynomials.
P(x) &= 2x^4-x^3+3x+4 I(x) &= x^2-4x+5
To find the voltage, these two polynomials have to be divided. Such division can be done following the steps of polynomial long division. First, write both polynomials in standard form, filling in any missing terms in the numerator with zeros. In P(x), only the quadratic term is missing.
P(x) &= 2x^4-x^3+ 0x^2+3x+4 I(x) &= x^2-4x+5
Now, write the dividend P(x) under the division symbol and the divisor I(x) to the left.
x^2-4x+5 & |r 2x^4-x^3+ 0x^2+3x+4
The first term of the quotient is found by dividing the leading term of P(x) by the leading term of I(x).
2x^4/x^2 = 2x^2
Next, write the monomial above the horizontal bar.
rr & 2x^2 x^2-4x+5 & |r 2x^4-x^3+0x^2+3x+4
Multiply the divisor by 2x^2 and write the resulting expression below the dividend. In this case, the multiplication gives 2x^4-8x^3+10x^2.
rr & 2x^2 x^2-4x+5 & |r 2x^4-x^3+0x^2+3x+4 & 2x^4-8x^3+10x^2
Subtract the resulting polynomial from the dividend.
rr & 2x^2 x^2 - 4x + 5 & |r 2x^4-x^3+0x^2+3x+4 & -( 2x^4-8x^3+10x^2) & 7x^3-10x^2+3x+4
Since the remainder has degree 3 and the divisor has degree 2, the division is not done yet. Therefore, the process will be repeated until the remainder's degree is less than 2.
The division is complete. The polynomial above the line is the quotient and the polynomial at the very bottom is the remainder. Q(x) &= 2x^2+7x+18 R(x) &= 40x-86 Finally, the voltage of the given circuit can be represented by the following expression.
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Discussion
Dividing a Polynomial by (x-k)
The polynomial long division is a powerful method that helps dividing any two given polynomials. However, when the divisor is a binomial of the form x-k, there is a shortcut that can be applied.
Method
Synthetic Division
When dividing a polynomial by a linear binomial, a binomial of the form x-k, there is an alternative method to the polynomial long division called the synthetic division. This method uses the constant term of the binomial and the coefficients of the numerator to compute the quotient. Consider the following division. x^3-3x^2+7/x-2 This division can be found following the next six steps.
1
Set Up the Division Using k and the Coefficients of the Numerator
expand_more Recall that the denominator is a linear binomial in the form x-k. The denominator is x-2, so the value of k is 2. The division symbol for synthetic division is L-shaped. The number k is written to the left.
rl r 2 & |cc & & &
After the numerator is written in standard form, its coefficients and constant term are written to the right of the division symbol. Fill in any missing terms with a zero.
rl r 2 & |cc 1 & -3 & 0 & 7
The numerator does not have a linear term, so a 0 was added between -3 and 7. Note that the number at the left of the division symbol is the opposite to the constant term of the divisor.
2
Bring Down the Leading Coefficient
expand_more Bring the leftmost coefficient down across the line. In this case, 1 is written below the horizontal line.
rl r 2 & |cc 1 & -3 & 0 & 7 ↓ & & & & c 1
3
Multiply k by the New Number Below the Line
expand_more Multiply the number written under the line in the previous step by k and write the product below the next coefficient above the line. In this case, 1 is multiplied by 2. The product 2 is written in the next column below -3.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & & & c 1
4
Add the Numbers in the Column Above the Line
expand_more Now, add the numbers in the column above the horizontal line, then write the result below the horizontal line in the same column. In this case, -3 and 2 are added and their sum, -1, is written below them and under the horizontal line.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & & & r 1 & -1
5
Repeat Steps 3 and 4 Through the Last Column
expand_more Steps 3 and 4 are repeated through the last column. In this exercise, -1 is multiplied by 2 and the product is written in the next column and above the line.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & -2 & & r 1 & -1 &
The numbers in this column are added and the sum is written below the line.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & -2 & & r 1 & -1 & -2
Then -2 is multiplied by 2 and the product is written in the next column and above the line.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & -2 & -4 & r 1 & -1 & -2
Finally, the numbers in the last column are added together and the sum is written below the line.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & -2 & -4 & r 1 & -1 & -2 & 3
The division is now complete.
6
Write the Quotient and Remainder
expand_more The rightmost number below the line is the remainder of the division. The remaining numbers below the line represent the coefficients of the quotient.
rl r 2 & |rr 1 & -3 & 0 & 7 & 2 & -2 & -4 & r 1 & -1 & -2 & 3
In this case, the remainder is 3. The quotient is a quadratic polynomial with coefficients 1, -1, and -2.
Q(x) &= x^2 - x - 2 r &= 3
Notice that the quotient is always one degree lower than the numerator because the denominator is a linear binomial. For the same reason, the remainder is always a number.
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Example
Using Synthetic Division
Consider the following polynomial. P(x) = 3x^4-4x^2-36
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b
Find the quotient and remainder of the division P(x)÷ (x-4).
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Hint
a
Use synthetic division. Remember to fill in the missing terms with zeros.
b
Use synthetic division. Remember to fill in the missing terms with zeros.
Solution
a
Since the divisor is a binomial of the form x-k, synthetic division can be used. First, note that P(x) has neither a cubic term nor a linear term. Therefore, fill in these two terms with zeros.
3x^4+0x^3-4x^2+0x-36/x+2 Next, set up the coefficients of P(x) and the synthetic division symbol. Here, k=-2. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 Bring down the first coefficient, multiply it by -2, and write the result in the second column. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 & -6 & c 3 Add the numbers in the second column and write the result below the line. Then multiply the new number below the line by -2 and write the result in the third column. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 & -6 & 12 & c 3 & -6 Next, add the numbers in the third column and write the result below the line. Multiply this number by -2 and write the result in the fourth column. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 & -6 & 12 & -16 & c 3 & -6 & 8 It is almost done! Add the numbers in the fourth column and write the result below the line. Then, multiply this number by -2 and write the result in the fifth column. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 & -6 & 12 & -16 & 32 & c 3 & -6 & 8 & -16 Finally, add the numbers in the last column and write the result below the line. rl r -2 & |rr 3 & 0 & -4 & 0 & -36 & -6 & 12 & -16 & 32 & c 3 & -6 & 8 & -16 & -4 The quotient of the division is formed by using the first four numbers below the line. The remainder is the very last number below the line. The quotient is one degree lower than the numerator. Q(x) &= 3x^3-6x^2+8x-16 r &= -4
b
As in Part A, the denominator is a linear binomial, x-4. For that reason, synthetic division can be used again. The coefficients at the right of the division symbol are the same, but this time the value of k is 4.
rl r 4 & |rr 3 & 0 & -4 & 0 & -36 The procedure to perform the division is similar to the one applied in Part A. First, bring the number 3 down. Next, multiply 4 by 3 and write the result in the second column and below 0. rl r 4 & |rr 3 & 0 & -4 & 0 & -36 & 12 & c 3 Next, find 0+12, which gives 12, and write the result in the same column but below the line. After that, multiply 4 by 12 and write the result in the third column and below -4. rl r 4 & |rr 3 & 0 & -4 & 0 & -36 & 12 & 48 & c 3 & 12 As in the previous step, add the numbers in the third column. The sum of -4 and 48 is 44, so write 44 in the same column but below the line. Then multiply -4 by 44 and write the resulting number in the fourth column and below 0. rl r 4 & |rr 3 & 0 & -4 & 0 & -36 & 12 & 48 & 176 & c 3 & 12 & 44 One more time, add the numbers in the fourth column and write the result in the same column but below the line. In this case, the sum is 176. Then, multiply 4 by 176 and write the result in the fifth column and below -36. rl r 4 & |rr 3 & 0 & -4 & 0 & -36 & 12 & 48 & 176 & 704 & c 3 & 12 & 44 & 176 Finally, find the sum of the numbers in the last column and write the result below the line. rl r 4 & |rr 3 & 0 & -4 & 0 & -36 & 12 & 48 & 176 & 704 & c 3 & 12 & 44 & 176 & 668 The first four numbers below the line are the coefficients of the quotient and the fifth number is the remainder. Q(x) &= 3x^3+12x^2+44x+176 r &= 668
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Discussion
The Remainder of a Polynomial Division
From synthetic division, the division P(x)÷ (x-k) can be written as the quotient Q(x) plus and the division of the remainder r and x-k. Here, r is a real number. P(x)/x-k = Q(x) + r/x-k However, this relation can be further manipulated. For example, the fractions can be eliminated entirely by multiplying both sides of the equation by x-k. P(x) =(x-k)Q(x) + r Two important results follow from the last equation.
Rule
The Remainder Theorem
If a polynomial P(x) is divided by a binomial of the form x-k, then the remainder of the division is equal to P(k).
r=P(k)
Proof
Consider the division of P(x) and a binomial of the form x-k.
P(x)/x-k
The division can be written in terms of the quotient and the remainder by synthetic division or polynomial long division.
Dividend/Divisor = Quotient + Remainder/Divisor ⇕ P(x)/x-k = Q(x) + R(x)/x-k
Since the divisor has degree 1 and the degree of the remainder has to be lower, the remainder has degree 0. This implies that R(x) is a constant. Thus, write r instead of R(x).
P(x)/x-k = Q(x) + r/x-k
Now, multiply both sides of the last equation by x-k.
P(x)/x-k = Q(x) + R(x)/x-k
MultEqn
LHS * (x-k)=RHS* (x-k)
P(x)=(Q(x)+r/x-k)(x-k)
Distr
Distribute (x-k)
P(x)=Q(x)(x-k) + r
Finally, evaluate the last equation at x=k.
P(x)=Q(x)(x-k) + r
Substitute
x= k
P( k)=Q( k)( k-k) + r
SubTerm
Subtract term
P(k)=Q(k)* 0 + r
ZeroPropMult
Zero Property of Multiplication
P(k) = r
RearrangeEqn
Rearrange equation
r = P(k)
Note that the Remainder Theorem gives the remainder of the division without the need to performing the division. For example, let P(x)=x^3-2x^2 +3x-4. To find the remainder of (x^3-2x^2 +3x-4)÷ (x-1), simply evaluate the numerator at 1. r = 1^3-2( 1)^2+3( 1)-4 ⇓ r = -2 Conversely, this theorem also says that the process of synthetic division can be used to evaluate a polynomial function at a given value. For example, to find P(-3), synthetic division can be used. rl r -3 & |rr 1 & -2 & 3 & -4 & -3 & 15 & -54 & r 1 & -5 & 18 & -58
From the above, P(-3)=-58.
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Discussion
Remainder 0
As when dividing integers, there is special significance when the remainder of a polynomial division is zero. First, it implies that the denominator divides the numerator perfectly, and, consequently, that the result of the division is a polynomial. P(x)/D(x) &= Q(x) + 0/D(x) &⇓ P(x) &= D(x) Q(x) Additionally, if the divisor is a linear binomial x-k, the fact that the remainder is 0 gives a connection between the denominator and the roots of the numerator. The following theorem expands on this concept.
Rule
The Factor Theorem
Let P(x) be a polynomial and k a real number. The binomial x-k is a factor of P(x) if and only if P(k)=0.
The binomial x-k is a factor of P(x) if and only if P(k)=0.
Notice that P(k)=0 means that k is a zero of P(x). Therefore, the theorem can also be stated as follows.
The binomial x-k is a factor of P(x) if and only if k is a zero of P(x).
The Factor Theorem is a special case of the Remainder Theorem and establishes a connection between the zeros of a polynomial and its factors.
Proof
Since the theorem is a biconditional statement, the proof will consists of two parts.
- Part I: If x-k is a factor of P(x), then P(k)=0.
- Part II: If P(k)=0, then x-k is a factor of P(x).
Part I
If x-k is a factor of P(x), then P(x) can be written as the product of x-k and a certain polynomial Q(x). P(x) = (x-k)Q(x) Now, evaluate the equation at x=k.
P(x) = (x-k)Q(x)
Substitute
x= k
P( k) = ( k-k)Q( k)
SubTerm
Subtract term
P(k) = 0* Q(k)
ZeroPropMult
Zero Property of Multiplication
P(k) = 0
Consequently, k is a zero of P(x). This completes the proof of the first part.
Part II
Consider the division of P(x) and x-k. P(x)/x-k The division can be rewritten in terms of the quotient and the remainder by using polynomial long division. P(x)/x-k = Q(x) + r/x-k By the Remainder Theorem, the remainder of the previous division is equal to P(k). Since P(k)=0, the remainder of the division is 0. Therefore, the rightmost term of the previous equation is 0. P(x)/x-k = Q(x) Finally, multiply both sides of the equation by x-k. P(x) = (x-k)Q(x) Consequently, x-k is a factor of P(x). This completes the proof of the second part.
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Example
Applying The Remainder Theorem
a
What is the remainder when P(x)=2x^5-3x^2+5x-10 is divided by x-3?
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b
Consider the polynomial H(x)=2x^5-13x^4+40x^2+2x-156. Without evaluating the polynomial, find H(6).
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Hint
a
b
Use the process of synthetic division to find H(6).
Solution
a
According to the Remainder Theorem, the remainder when a polynomial P(x) is divided by a binomial of the form x-k is equal to the polynomial evaluated at x=k.
r = P(k) In the given case, the binomial is x-3, which means that k=3. Therefore, instead of performing the division, the remainder can be found simply by evaluating P(x) at x=3.
P(x) = 2x^5-3x^2+5x-10
▼
Substitute 3 for x and evaluate
Substitute
x= 3
P( 3) = 2( 3)^5-3( 3)^2+5( 3)-10
CalcPowProd
Calculate power and product
P(3) = 2(243)-3(9)+15-10
Multiply
Multiply
P(3) = 486-27+15-10
AddSubTerms
Add and subtract terms
P(3) = 464
Consequently, the remainder is 464 — that is, r=464. To verify that the result is correct, perform the polynomial long division.
As shown, the remainder of the division is 464, as previously concluded. The remainder could also be found using synthetic division.
b
Synthetic division can be used to find H(6) without evaluating the polynomial. Rather, the process of synthetic substitution will be used. First, set up the division symbol and the coefficients of P(x). Remember to fill in the missing terms with zeros. Then, write 6 to the left.
rl r 6 & |cc 2 & -13 & 0 & 40 & 2 & -156 Next, follow the steps of synthetic division. Bring down the number 2. Multiply 6 by 2 and write the result below -13. rl r 6 & |rr 2 & -13 & 0 & 40 & 2 & -156 & 12 & & & c 2 Add the numbers in the second column and write the result below the line. rl r 6 & |rr 2 & -13 & 0 & 40 & 2 & -156 & 12 & & & c 2 & -1 The same steps are repeated over and over until the last column is reached. rl r 6 & |rr 2 & -13 & 0 & 40 & 2 & -156 & 12 & -6 & -36 & 24 & 156 & c 2 & -1 & -6 & 4 & 26 & At the end of the process, it was concluded that H(6)=0. Therefore, 6 is a zero of P(x). Consequently, by the Factor Theorem, x-6 is a factor of H(x). In fact, H(x) can be written as follows. 2x^5-13x^4+40x^2+2x-156 = (x-6)(2x^4-x^3-6x^2+4x+26)
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Pop Quiz
Finding Remainders
Find the remainder of the given polynomial division.
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Closure
Comparing Polynomial Long Division and Synthetic Division
When it comes to dividing polynomials, there are two main methods that can be applied — namely, polynomial long division and synthetic division. The table below lists some pros and cons of each method.
| Pros | Cons | |
|---|---|---|
| Polynomial Long Division | Works for every polynomial division | Involves variables, powers, and many computations |
| Synthetic Division | Involves only numbers | Works only when the divisor has the form x-k |
Keep in mind that synthetic division is a shortcut of polynomial long division. In fact, synthetic division can be derived from the process of polynomial long division. Consider, for example, the following division. 2x^5-3x^4+2x^2+6/x-3 This calculation can be solved by using long division.
To clarify or simplify the calculations, the minus signs can be removed by distributing them to the parentheses. Also, the remaining terms in the dividend can be hidden in the third, fifth, and seventh rows until the terms are needed. This cleaner look can be seen below.
Now, take a look at the process for computing the same division using synthetic division. rl r 3 & |rr 2 & -3 & 0 & 2 & 0 & 6 & 6 & 9 & 27 & 87 & 261 & r 2 & 3 & 9 & 29 & 87 & 267 Notice that the numbers below the horizontal line are the leading coefficients of the polynomials in the first, third, fifth, seventh, and ninth rows of the polynomial long division. Furthermore, the additions performed in synthetic division can also be located in the long division process.
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Dividing Polynomials
Exercise 2.1
Which of the following binomials are factors of P(x)=x^4 - 2x^3 - 33x^2 - 22x + 56?
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Instead of factoring the given polynomial, we can use the Factor Theorem, which states that a binomial x-k is a factor of a polynomial P(x) if and only if P(k)=0. In other words, we just need to evaluate the polynomial at the corresponding value and see whether the result is 0.
| Binomial | Value of k |
|---|---|
| x-1 | 1 |
| x-2 | 2 |
| x+4 | -4 |
| x+1 | -1 |
To determine which of the given binomials are factors of P(x), we will evaluate P(x) at the values of k written in the right-hand column. Let's begin with k=1.
Since P(1)=0, by the Factor Theorem, the binomial x-1 is a factor of P(x). In a similar way, we evaluate the remaining values of k.
| Binomial | Value of k | P(k) | Result | Is It a Factor? |
|---|---|---|---|---|
| x-2 | 2 | P( 2) = 2^4 - 2( 2)^3 - 33( 2)^2 - 22( 2) + 56 | -120 | No |
| x+4 | -4 | P( -4) = -4^4 - 2( -4)^3 - 33( -4)^2 - 22( -4) + 56 | 0 | Yes |
| x+1 | -1 | P( -1) = -1^4 - 2( -1)^3 - 33( -1)^2 - 22( -1) + 56 | 48 | No |
We can use this information to conclude that, of the given binomials, only x-1 and x+4 are factors of P(x).
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We begin by noticing that the divisor is a binomial of the form x-k. This allows us to use the Remainder Theorem.
Remainder Theorem |- If a polynomial P(x) is divided by a binomial of the form x-k, then the remainder of the division is equal to P(k).
In the question, it is a given that the remainder of f(x)÷(x-k) is equal to -5. This implies that f(k)=-5. To determine the value of k, in the given graph, let's check the value of x for which the output is -5.
As seen, when x= 3, we get that f( 3)= -5. Therefore, the value of k is 3.
As in Part A, we are interested in the values of x for which the output of f(x) is equal to 2. To do so, let's check the given graph.
From the graph, we can see that f(x) is equal to 2 for x= -4, 0, 1. Therefore, the remainder of f(x)÷ (x-t) is 2 for t=-4, 0, and 1.
We can apply the same reasoning as in the previous parts to determine the value of p for which the remainder of f(x)÷(x-p) is equal to 1, let's check when the graph is equal to 1.
There are three values of x for which f(x) is equal to 1. These values are -3.64, -1, and 1.64. However, because the value of p we are looking for is an integer, we discard -3.64 and 1.64. Therefore, the value of p is -1.
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To find the value of g such that the given division has a remainder of zero, we can use the Remainder Theorem.
Remainder Theorem |- If a polynomial P(x) is divided by a binomial of the form x-k, then the remainder of the division is equal to P(k).
Since the denominator is x-2, we will substitute x=2 into the numerator and set it equal to 0. ( 2)^3 - 6( 2)^2 + g( 2) + 42 = 0 Next, let's solve the equation!
Consequently, the value of g for which the given division has remainder 0 is -13.
We can find the value of g by performing the given division using synthetic division.
rl IR-0.15cm r 2 & |rr 1 & -6 & g & 42 & 2 & -8 & 2g - 16 & c 1 & -4 & g - 8 & 2g + 26
We are told that the remainder of the division is equal to 0. Therefore, 2g+26=0. Solving this equation gives us the value of g.
2g+26 = 0 ⇒ g = -13
Following a similar procedure as in Part A, we will evaluate the numerator at x=2 and equate it to 40. ( 2)^3 - 6( 2)^2 + g( 2) + 42 = 40 As before, let's solve the equation!
In this case, the value of g=7.
We can find the value of g by performing the given division using synthetic division.
rl IR-0.15cm r 2 & |rr 1 & -6 & g & 42 & 2 & -8 & 2g - 16 & c 1 & -4 & g - 8 & 2g + 26
We know that the remainder of the division is equal to 40. This means that 2g+26=40. Solving this equation gives us the value of g.
2g+26 = 40 ⇒ g = 7
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If a polynomial of degree 6 is divided by a polynomial of degree 4, what is the degree of the quotient?
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We begin by remembering that the numerator can be written as the quotient multiplied by the divisor plus the remainder. P(x) = Q(x)D(x) + R(x) From this equation, we know that the degree of the right-hand side polynomial must be equal to the degree of P(x). In this case, it is a given that the degree of P(x) is 6. Thus, the degree of the right-hand side must be 6. P(x)_(Degree: 6) = Q(x)D(x) + R(x)_(Degree: 6) Let q be the degree of the quotient. From the polynomial long division, we know that the degree of the remainder is less than the degree of the divisor. Since the divisor has degree 4, the remainder has degree 3 or lower.
When two polynomials are added, the degree of the resulting polynomial is equal to the degree of the polynomial of higher degree. The degree ofR(x)does not affect the degree of the right-hand side. Additionally, when two polynomials are multiplied, the degree of the resulting polynomial is equal to the sum of the degrees of the polynomials being multiplied. Therefore, using the degrees of Q(x) and D(x), we can set and solve an equation for q. 6 = q+4 ⇒ q = 2 Consequently, the degree of the quotient is 2.
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Dividing Polynomials
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