Detailed instructions for solving Problem 7.44
* How to add (subtract) 2 polynomials:
Method 1: Remove the brackets and group the terms of the same degree.
Method 2: Set the addition (subtract) so that the terms of the same degree are placed in the same column and then add (subtract) by each column.
* How to multiply 2 polynomials:
Method 1: To multiply a polynomial by a polynomial, we multiply each term of this polynomial by each term of the other polynomial and then add the products together.
Method 2: Set the multiplier:
+ Multiply each term in the lower row by the polynomial in the upper line and write the result in a separate line.
+ Write the lines so that the terms of the same order are in the same column to perform the addition by column.
* To divide polynomial A by polynomial B, we do the following:
Step 1: Set the divisibility similar to dividing two natural numbers. Divide the highest-order term of A by the highest-order term of B.
Step 2: Take A minus the product of B with the new quotient obtained in step 1
Step 3: Divide the highest order term of the first remainder by the highest term of B
Step 4: Subtract the product B from the first remainder with the quotient obtained in step 3
Step 5: Do the same as above
When the last remainder is of degree less than the degree of B, the division ends.
a) We have:
B = (A + B) – A
= (x3 + 3x + 1) – (x4 + x3 – 2x – 2)
= x3 + 3x + 1 – x4 – x3 + 2x + 2
= – x4 + (x3 – x3) + (3x + 2x) + (1 + 2)
= – x4 + 5x + 3
b) C = (A – C) – A
= x5 – (x4 + x3 – 2x – 2)
= x5 – x4 – x3 + 2x + 2)
c) D = (2x3 – 3) . A
= (2x3 – 3) . (x4 + x3 – 2x – 2)
= 2x3 . (x4 + x3 – 2x – 2) + (-3) .(x .)4 + x3 – 2x – 2)
= 2x3 . x4 + 2x3 . x3 + 2x3 . (-2x) + 2x3 . (-2) + (-3). x4 + (-3) . x3 + (-3). (-2x) + (-3). (-2)
= 2x7 + 2x6 – 4x4 – 4x3 – 3x4 – 3x3 + 6x + 6
= 2x7 + 2x6 + (-4x4 – 3x4) + (-4x3 – 3x3) + 6x + 6
= 2x7 + 2x6 – 7x4 – 7x3 + 6x + 6
d) P = A : (x+1) = (x4 + x3 – 2x – 2) : (x + 1)
So P = x3 – 2
e) Q = A : (x2 + 1)
If A divided by the polynomial x2 + 1 has no remainder, then there is a satisfying polynomial Q
We perform division (x4 + x3 – 2x – 2) : (x2 + 1)
Since division has remainders, there is no satisfying polynomial Q
Solution 7.45 page 46 Math textbook 7 Connecting knowledge volume 2 – KNTT
Let the polynomial P(x). Explain why if there is a polynomial Q(x) such that P(x) = (x – 3). Q(x) (i.e. P(x) is divisible by x – 3) then x = 3 is a solution of P(x)
Detailed instructions for solving Problem 7.45
The root of the polynomial variable x is the value of x where the polynomial has a value of 0.
Because at x = 3 then P(x) = (3 – 3) . Q(x) = 0. Q(x) = 0 so x = 3 is a solution of the polynomial P(x)
Solution 7.46 page 46 Math textbook 7 Connecting knowledge volume 2 – KNTT
Two friends Round and Square argue with each other as follows:
Please tell me your opinion and give an illustrative example.
Detailed instructions for solving problems 7.46
The sum of polynomials is a polynomial of degree no greater than the degree of the component polynomials
Round is true, Square is false because the sum of the polynomials is a polynomial of degree not greater than the degree of the component polynomials.
Polynomial M(x) = x3 + 1 can be written as the sum of two quadratic polynomials whose highest coefficients are two opposite numbers.
x3 + 1 = (x4 + 1) + (-x4 + x3)