Algebraic Identities

Algebraic identities are fundamental equations that hold true for any value of their variables.

This section introduces several important algebraic identities, including the squares of binomials and cubic formulas. It also illustrates their applications in expanding, factoring, and calculating products using the identities.

Algebraic Identities

Algebraic identities are equations that remain true regardless of the values of their variables. This section reviews key identities that include:

• Identity I: $(x + y)^2 = x^2 + 2xy + y^2$
• Identity II: $(x - y)^2 = x^2 - 2xy + y^2$
• Identity III: $x^2 - y^2 = (x + y)(x - y)$
• Identity IV: $(x + a)(x + b) = x^2 + (a + b)x + ab$

The section includes examples demonstrating how to find products using these identities and emphasizes their utility for both expansion and factorization. Additionally, more complex identities such as those involving three variables and cubes of binomials are introduced, expanding the toolbox of algebraic techniques.

Example

Factorise:
Let
$p(x) = x^3 - 25x^2 + 156x - 150.$

We will also look for all the factors of $-150$. Some of these are $\\pm 1, \\pm 2, \\pm 3, \\pm 5, \\pm 6, \\pm 10, \\pm 15, \\pm 25, \\pm 30, \\pm 50, \\pm 75, \\pm 150$.

By trial, we find that $p(1) = 0$. So, $1$ is a factor of $p(x)$.

Now we see that $x^3 - 25x^2 + 156x - 150 = (x - 1)(x^2 - 24x + 150)$ $(Why?)$

We could have also got this by dividing $p(x)$ by $(x - 1)$ using the Factor theorem. By splitting the middle term, we have:

$\ x^2 - 24x + 150 = (x - 10)(x - 15)$

So,
$p(x) = (x - 1)(x - 10)(x - 15)$

Similar Question:

Factorise:
Let
$p(x) = x^3 - 30x^2 + 195x - 270.$

We will also look for all the factors of $-270$. Some of these are $\\pm 1, \\pm 2, \\pm 3, \\pm 5, \\pm 6, \\pm 9, \\pm 10, \\pm 15, \\pm 18, \\pm 27, \\pm 30, \\pm 45, \\pm 90, \\pm 135, \\pm 270$.

By trial, we find that $p(3) = 0$. So, $3$ is a factor of $p(x)$.

Now we see that $x^3 - 30x^2 + 195x - 270 = (x - 3)(x^2 - 27x + 90)$ $(Why?)$

We could have also got this by dividing $p(x)$ by $(x - 3)$ using the Factor theorem. By splitting the middle term, we have:

$\ x^2 - 27x + 90 = (x - 9)(x - 10)$

So,
$p(x) = (x - 3)(x - 9)(x - 10)$

Note: Ensure to verify all factor pairs and confirm through substitution to maintain solution integrity.