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In this chapter, we transition from linear polynomials to Algebraic Identities—special rules that simplify the most complex calculations.
Take any 3 consecutive square numbers (e.g., \( 25, 36, 49 \)). Add the smallest and largest, then subtract twice the middle:
Wait... the result is always 2! We will soon use algebra to prove why.
Imagine a line segment of length \( (a+b) \). A square constructed on this line can be partitioned into four distinct areas.
\( (a + b)^2 = a^2 + 2ab + b^2 \)
Click to see the rational number check (\( a = -2/3, b = 3/4 \)).
Equation: \( x^2-1=24 \) is true only for \( x = \pm 5 \).
Identity: \( (a+b)^2 \) is true for EVERY value.
Is \( (a+b)^2 > a^2 + b^2 \)? Yes, if \( a, b \) are positive, because of the extra \( 2ab \) term!
Identities can be reversed to find factors. For example, \( x^2 + 4x + 4 = (x+2)^2 \).
James expands first, while Reshma groups terms as \( (a-b)[(a-b)(a+b)] = (a-b)(a^2-b^2) \). Both are correct, but Reshma's path is often faster!
Subtract the rectangles from a large square of side \( a \):
What if we square three numbers? \( (a+b+c)^2 \)
Visualise \( x^2+7x+12 \) as a rectangle with sides \( (x+3) \) and \( (x+4) \).
Saira has \( x^2 + 8x + 15 \). By splitting 8x into 3x and 5x, she finds the dimensions: \( (x+3) \) by \( (x+5) \).
Expanding into the 3rd dimension: \( (a+b)^3 \) can be split into 2 cubes and 6 cuboids.
If \( x+y+z=10, xyz=25, x^2+y^2+z^2=38 \), then \( x^3+y^3+z^3 = \mathbf{145} \)!
We simplify rational expressions by cancelling non-zero common factors.
\(\frac{x^2 - 7x + 12}{5x^2 + 5x - 100} = \frac{(x-3)(x-4)}{5(x-4)(x+5)} = \frac{x-3}{5(x+5)}\)