যেভাবে দুই বা ততোধিক অজ্ঞাত রাশির সাথে বহুপদীসমূহকে গুণ করতে হয়

We know how to add and subtract polynomials with multiple unknowns, we know how to multiply two binomials together (FOIL), and we know how to multiply a monomial by a polynomial (basic distributive property).

But what are we supposed to do if the expressions that we're multiplying together have more than one variable (like $x$ and $y$ instead of just $x$)?

Not to worry! We're actually going to use the same approaches as we always have:

• If we're multiplying two binomials together, we'll use FOIL.
• For all other kinds of polynomial multiplication, we're still going to apply the basic distributive property.

Right—when we're dealing with a pair of binomials, the FOIL method reminds us which of those polynomials' terms should be multiplied together:

F: First two terms (in red below)

O: Outside two terms (in orange below)

I: Inside two terms (in yellow below)

L: Last two terms (in green below)

Let's look at some examples.

Example 1

$$(3x-4y)(2x+5y)$$

We've got two binomials here, so let's use FOIL.

$$=6x^2 +15xy -8xy -20y^2$$ $$=6x^2 +7xy -20y^2$$

Notice how we still used FOIL despite having two binomials that contain multiple variables. Remember your rules for adding monomials - do not add the exponents!

So, when we had to multiply a term containing an $x$ variable and a term containing a $y$ variable together, we got a term with an $xy$ (like $15xy$). Note that multiplying two $x$ terms together gives us the usual $x^2$, and that multiplying two $y$ terms together still gives us a $y^2$.

Example 2

$$(x+7y)(2x-3y+5)$$

We're going to apply the regular ol' distributive property here. This means that we need to distribute each term in the first set of parentheses to each term in the second set of parentheses. Think of it like dragging the terms through, multiplying them along the way.

All right, let's put this method into practice: give the example above a shot and then check your answer below!

Example 3

$$(2y-x)(3y+5x) + (-y+6x)(2y+3x)$$

Ugh. That looks horrible! But, when you take a deep breath, you'll notice that this expression is just two pairs of binomials being multiplied together.

So, let's use FOIL on both pairs of binomials and then add everything together.

(Video) Multiplying Multivariable Polynomials

by Krista King

This video by the Youtube channel CalculusExpert.com explains how to multiply polynomials with multiple variables.

$\large{\text{Summary}}$

Luckily, multiplying multivariable polynomials isn't different from multiplying regular polynomials. You might have more variables but you multiply the same.

The first example she looks at is: $$(2x+y)(x-2y)-(x+2y)(2x-y)$$

Let's start by multiplying the first two binomials together. Pay special attention to the red areas. This is where the different variables multiply together. $$\begin{align} &(2x+y)(x-2y)\\ &=(2x\cdot x)+\color{#29A603}{(2x\cdot-2y)}+\color{#29A603}{(y\cdot x)}+(y\cdot-2y)\\ &=2x^2-4xy+xy-2y^2\\ &=2x^2-3xy-2y^2 \end{align}$$

Now let's multiply the next two together. Again, the red shows where we multiply multiple variables. $$\begin{align} &(x+2y)(2x-y)\\ &=(x\cdot2x)+\color{#29A603}{(x\cdot-y)}+\color{#29A603}{(2y\cdot2x)}+(2y\cdot-y)\\ &=2x^2-xy+4xy-2y^2\\ &=2x^2+3xy-2y^2 \end{align}$$

Now, we want to subtract these two polynomials. $$\begin{align} &(2x+y)(x-2y)-(x+2y)(2x-y)\\ &=(2x^2-3xy-2y^2)-(2x^2+3xy-2y^2)\\ &=2x^2-3xy-2y^2-2x^2-3xy+2y^2\\ &=(2x^2-2x^2)+(-3xy-3xy)+(-2y^2+2y^2)\\ &=-6xy \end{align}$$

We use the FOIL method when multiplying two binomials, but when at least one of the polynomials we're multiplying together has more than two terms, we cannot use the FOIL method.

Instead, we multiply each term in the first polynomial by each term in the second polynomial.

Let's try an example together: $$(x+a)(2x+3y-5)$$ First, we multiply $x$ by each term in the second polynomial: $$x(2x+3y-5)=2x^2+3xy-5x$$ Then, we multiply $a$ by each term in the second polynomial: $$a(2x+3y-5)=2ax+3ay-5a$$ Finally, we add up all of the terms that resulted from the multiplication we just did: $$(x+a)(2x+3y-5) = 2x^2+3xy-5x+2ax+3ay-5a$$

Now that you've got the hang of it, let's try some additional examples to help solidify what you've just learned:

$$(x + 2y)(3x − 4y + 5)$$ $$= x(3x − 4y + 5) + 2y(3x − 4y + 5)$$ $$= 3x^2 − 4xy + 5x + 6xy − 8y^2 + 10y$$

Finally, we combine like terms: $$3x^2 − 4xy + 5x + 6xy − 8y^2 + 10y$$ $$= 3x^2 + 2xy + 5x − 8y^2 + 10y$$

$$(3x - 4y)(-2x + 5y - 6)$$ $$= 3x(-2x + 5y - 6) - 4y(-2x + 5y - 6)$$ $$= -6x^2 + 15xy - 18x + 8xy - 20y^2 + 24y$$ $$= -6x^2 + 23xy - 18x - 20y^2 + 24y$$

$$(3a - 4b + 1)(2a + 5b - 3)$$ $$= 3a(2a + 5b - 3) - 4b(2a + 5b - 3) + 1(2a + 5b - 3)$$ $$= 6a^2 + 15ab - 9a - 8ab - 20b^2 + 12b + 2a + 5b - 3$$ $$= 6a^2 - 20b^2 + 7ab - 7a + 17b - 3$$

Multiply Polynomials: $2+$ Unknowns

We've covered how to multiply polynomials from all angles...except when we have multiply unknowns!

Don't fret. You will use almost the same technique

• Distribute
• Simplify (if necessary)

When you're multiply polynomials with multiple unknowns, always be mindful of your variables! For example, as we can see in the image, when we multiply $2x*3y$, our new monomial must have both variables, $6xy$.