Ontario 10 Applied (MFM2P)

Components in an expression

Lesson

We've already learnt that, in algebra, we use variables to represent unknown values. A run down of of the basic components of an algebraic expression can be found here.

Once we understand all these components, we can use them to write more complex relationships between different variables, though there is one more we need to introduce first.

We've worked with the highest numeric and algebraic common factors of expressions previously, but now we want to consider all possible factors, not just the highest.

Remember!

We can break an expression down into a product of smaller parts. If these parts happen to be integers or algebraic expressions, we call them *factors*. For example, consider the following expressions:

$2x$2`x`

- In how many ways can we break this expression down? There are two unique ways to write this as a product of integers and algebraic terms: $1\times2x$1×2
`x`and $2\times x$2×`x`. This means our factors are $1$1, $2x$2`x`, $2$2 and $x$`x`.

$16x$16`x`

- There are in fact five distinct ways to write this as a product. $1\times16x$1×16
`x`, $2\times8x$2×8`x`,$4\times4x$4×4`x`, $16\times x$16×`x`, and $8\times2x$8×2`x`. So we have ten factors: $1$1, $2$2, $4$4, $8$8, $16$16, $x$`x`, $2x$2`x`, $4x$4`x`, $8x$8`x`and $16x$16`x`.

$3\left(x+3\right)$3(`x`+3)

- There are two different ways to write this as a product. $1\times3\left(x+3\right)$1×3(
`x`+3) and $3\times\left(x+3\right)$3×(`x`+3). So our factors are $1$1, $3$3, $x+3$`x`+3 and $3\left(x+3\right)$3(`x`+3).

$2x+2$2`x`+2

- There are also two ways to write this as a product. $1\times\left(2x+2\right)$1×(2
`x`+2) and $2\times\left(x+1\right)$2×(`x`+1). So our factors are $1$1, $2$2, $x+1$`x`+1 and $2x+2$2`x`+2.

Notice that $1$1 and the expression itself are always factors.

Now let's look at some more examples to further explore the structure of expressions.

How do $4x^4$4`x`4 and $6x^2$6`x`2 relate to the expression $4x^4+6x^2$4`x`4+6`x`2? Choose the correct answer from the options below.

$4x^4$4

`x`4 and $6x^2$6`x`2 are factors of $4x^4+6x^2$4`x`4+6`x`2.A$4x^4$4

`x`4 and $6x^2$6`x`2 are terms of $4x^4+6x^2$4`x`4+6`x`2.B$4x^4$4

`x`4 and $6x^2$6`x`2 are coefficients of $4x^4+6x^2$4`x`4+6`x`2.C$4x^4$4

`x`4 and $6x^2$6`x`2 are multiples of $4x^4+6x^2$4`x`4+6`x`2.D$4x^4$4

`x`4 and $6x^2$6`x`2 are factors of $4x^4+6x^2$4`x`4+6`x`2.A$4x^4$4

`x`4 and $6x^2$6`x`2 are terms of $4x^4+6x^2$4`x`4+6`x`2.B$4x^4$4

`x`4 and $6x^2$6`x`2 are coefficients of $4x^4+6x^2$4`x`4+6`x`2.C$4x^4$4

`x`4 and $6x^2$6`x`2 are multiples of $4x^4+6x^2$4`x`4+6`x`2.D

What is the highest constant factor of the following expression?

$4x+12$4`x`+12

Which of the options are factors of the expression below? Select all that apply.

$9\left(x+2\right)^2$9(

`x`+2)2$x$

`x`A$9$9

B$9\left(x+2\right)^2$9(

`x`+2)2C$2$2

D$x+2$

`x`+2E$\left(x+2\right)^2$(

`x`+2)2F$x$

`x`A$9$9

B$9\left(x+2\right)^2$9(

`x`+2)2C$2$2

D$x+2$

`x`+2E$\left(x+2\right)^2$(

`x`+2)2F

Expand and simplify second-degree polynomial expressions involving one variable that consist of the product of two binomials [e.g., (2x + 3)(x + 4)] or the square of a binomial [e.g., (x + 3)^2], using a variety of tools and strategies