## Detailed instructions for solving Problem 7.44

**Solution method**

* 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.

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+ 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.

**Detailed explanation**

a) We have:

B = (A + B) – A

= (x^{3} + 3x + 1) – (x^{4} + x^{3} – 2x – 2)

= x^{3} + 3x + 1 – x^{4} – x^{3} + 2x + 2

= – x^{4} + (x^{3} – x^{3}) + (3x + 2x) + (1 + 2)

= – x^{4} + 5x + 3

b) C = (A – C) – A

= x^{5} – (x^{4} + x^{3} – 2x – 2)

= x^{5} – x^{4} – x^{3} + 2x + 2)

c) D = (2x^{3} – 3) . A

= (2x^{3} – 3) . (x^{4} + x^{3} – 2x – 2)

= 2x^{3} . (x^{4} + x^{3} – 2x – 2) + (-3) .(x .)^{4} + x^{3} – 2x – 2)

= 2x^{3} . x^{4} + 2x^{3} . x^{3} + 2x^{3} . (-2x) + 2x^{3} . (-2) + (-3). x^{4} + (-3) . x^{3} + (-3). (-2x) + (-3). (-2)

= 2x^{7} + 2x^{6} – 4x^{4} – 4x^{3} – 3x^{4} – 3x^{3} + 6x + 6

= 2x^{7} + 2x^{6} + (-4x^{4} – 3x^{4}) + (-4x^{3} – 3x^{3}) + 6x + 6

= 2x^{7} + 2x^{6} – 7x^{4} – 7x^{3} + 6x + 6

d) P = A : (x+1) = (x^{4} + x^{3} – 2x – 2) : (x + 1)

So P = x^{3} – 2

e) Q = A : (x^{2} + 1)

If A divided by the polynomial x^{2} + 1 has no remainder, then there is a satisfying polynomial Q

We perform division (x^{4} + x^{3} – 2x – 2) : (x^{2} + 1)

Since division has remainders, there is no satisfying polynomial Q

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### 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

**Solution method**

The root of the polynomial variable x is the value of x where the polynomial has a value of 0.

**Detailed explanation**

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)

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### 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

**Solution method**

The sum of polynomials is a polynomial of degree no greater than the degree of the component polynomials

**Detailed explanation**

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) = x^{3} + 1 can be written as the sum of two quadratic polynomials whose highest coefficients are two opposite numbers.

For example:

x^{3} + 1 = (x^{4} + 1) + (-x^{4} + x^{3})

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