CONJUGATE BINOMIAL: HOW TO SOLVE IT, EXAMPLES, EXERCISES - MATHS - 2024

A conjugate binomial of another binomial is one in which they are only differentiated by a sign of the operation. The binomial, as its name implies, is an algebraic structure consisting of two terms.

Some examples of binomials are: (a + b), (3m - n) and (5x - y). And their respective conjugated binomials are: (a - b), (-3m - n) and (5x + y). As can be seen immediately, the difference is in the sign.

Figure 1. A binomial and its conjugate binomial. They have the same terms, but differ in sign. Source: F. Zapata.

A binomial multiplied by its conjugate results in a remarkable product that is widely used in algebra and science. The result of the multiplication is the subtraction of the squares of the terms of the original binomial.

For example, (x - y) is a binomial and its conjugate is (x + y). So, the product of the two binomials is the difference of the squares of the terms:

(x - y). (x + y) = x ² - y ²

How do you solve a conjugate binomial?

The stated rule of conjugated binomials is the following:

As an example of application, we will begin by demonstrating the previous result, which can be done using the distributive property of the product with respect to the algebraic sum.

(x - y) (x + y) = xx + xy - yx - yy

The above multiplication was obtained by following these steps:

- The first term of the first binomial is multiplied by the first term of the second

- Then the first of the first, for the second of the second

- Then the second of the first by the first of the second

- Finally the second of the first by the second of the second.

Now let's make a small change using the commutative property: yx = xy. It looks like this:

(x - y) (x + y) = xx + xy - xy - yy

As there are two equal terms but of opposite sign (highlighted in color and underlined), they are canceled and it is simplified:

(x - y) (x + y) = xx - yy

Finally, it is applied that multiplying a number by itself is equivalent to raising it to the square, so that xx = x ² and also yy = y ².

In this way it is demonstrated what had been indicated in the previous section, that the product of a sum and its difference is the difference of the squares:

(x - y). (x + y) = x ² - y ²

Figure 2. A sum times its difference is a difference of squares. Source: F. Zapata.

Examples

- Conjugated binomials of various expressions

Example 1

Find the conjugate of (y ² - 3y).

Answer: (y ² + 3y)

Example 2

Obtain the product of (y ² - 3y) and its conjugate.

Answer: (y ² - 3y) (y ² + 3y) = (y ²) ² - (3y) ² = y ⁴ - 3 ² y ² = y ⁴ - 9y ²

Example 3

Develop the product (1 + 2a). (2a -1).

Answer: the previous expression is equivalent to (2a + 1). (2a -1), that is, it corresponds to the product of a binomial and its conjugate.

It is known that the product of a binomial by its conjugate binomial is equal to the difference of the squares of the terms of the binomial:

(2a + 1) (2a -1) = (2a) ² - 1 ² = 4 a ² - 1

Example 4

Write the product (x + y + z) (x - y - z) as a difference of squares.

Answer: we can assimilate the above trinomials to the conjugate binomial form, making careful use of parentheses and square brackets:

(x + y + z) (x - y - z) =

In this way the difference of squares can be applied:

(x + y + z) (x - y - z) =. = x ² - (y + z) ²

Example 5

Express the product (m ² - m -1). (M ² + m -1) as a difference of squares.

Answer: the previous expression is the product of two trinomials. It must first be rewritten as the product of two conjugated binomials:

(m ² - m -1) (m ² + m -1) = (m ² - 1 - m) (m ² -1 + m) =.

We apply the fact that the product of a binomial by its conjugate is the quadratic difference of its terms, as has been explained:

. = (m ² -1) ² - m ²

Exercises

As always, you start with the simplest exercises and then increase the level of complexity.

- Exercise 1

Type (9 - to ²) as a product.

Solution

First, we rewrite the expression as a difference of squares, in order to apply what was previously explained. Thus:

(9 - a ²) = (3 ² - a ²)

Next we factor, which is equivalent to writing this difference of squares as a product, as requested in the statement:

(9 - a ²) = (3 ² - a ²) = (3 + a) (3 -a)

- Exercise 2

Factor 16x ² - 9y ⁴.

Solution

Factoring an expression means writing it as a product. In this case, it is necessary to previously rewrite the expression, to obtain a difference of squares.

It is not difficult to do this, since looking carefully, all the factors are perfect squares. For example 16 is the square of 4, 9 is the square of 3, and ⁴ is the square of y ² and x ² is the square of x:

16x ² - 9y ⁴ = 4 ² x ² - 3 ² y ⁴ = 4 ² x ² - 3 ² (y ²) ²

Then we apply what we already know previously: that a difference of squares is the product of conjugated binomials:

(4x) ² - (3 and ²) ² = (4x - 3 and ²). (4x + 3 and ²)

- Exercise 3

Write (a - b) as a product of binomials

Solution

The above difference should be written as differences of squares

(√a) ² - (√b) ²

Then it is applied that the difference of squares is the product of the conjugated binomials

(√a - √b) (√a + √b)

- Exercise 4

One of the uses of the conjugate binomial is the rationalization of algebraic expressions. This procedure consists of eliminating the roots of the denominator of a fractional expression, which in many cases facilitates the operations. It is requested to use the conjugate binomial to rationalize the following expression:

√ (2-x) /

Solution

The first thing is to identify the conjugate binomial of the denominator:.

Now we multiply the numerator and denominator of the original expression by the conjugate binomial:

√ (2-x) / {.}

In the denominator of the previous expression we recognize the product of a difference by a sum, which we already know corresponds to the difference of the squares of the binomials:

√ (2-x). / {(√3) ² - ² }

Simplifying the denominator is:

√ (2-x). / = √ (2-x). / (1 - x)

Now we deal with the numerator, for which we will apply the distributive property of the product with respect to the sum:

√ (2-x). / (1 - x) = √ (6-3x) + √ / (1 - x)

In the previous expression we recognize the product of the binomial (2-x) by its conjugate, which is the notable product equal to the difference of squares. In this way, a rationalized and simplified expression is finally obtained:

/ (1 - x)

- Exercise 5

Develop the following product, using the properties of the conjugate binomial:

Solution

4a ^{(2x + 6y)} - 9a ^{(2x - 6y)} = 4a ^(2x).a ^(6y) - 9a ^(2x).a ^(-6y) =.a ^(2x)

The attentive reader will have noticed the common factor that has been highlighted in color.

References

1. Baldor, A. 1991. Algebra. Editorial Cultural Venezolana SA
2. González J. Conjugated binomial exercises. Recovered from: academia.edu.
3. Math teacher Alex. Remarkable products. Recovered from youtube.com.
4. Math2me. Conjugated binomials / notable products. Recovered from youtube.com.
5. Conjugated binomial products. Recovered from: lms.colbachenlinea.mx.
6. Vitual. Conjugated binomials. Recovered from: youtube.com.