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# Lecture 17: October 27, 2023 ## Quick notes * No class on Nov 10 ## Factoring Polynomials ### Example 1 To factor $x^2 + bx + c$, find two numbers $r$ and $s$ such that $r + s = b$ and $rs = c$. $x^2 + 5x + 6 = (x + 3)(x + 2)$ $x^2 - 5x + 6 = (x - 3)(x - 2)$ $x^2 + 5x - 6 = (x + 6)(x - 1...

# Lecture 17: October 27, 2023 ## Quick notes * No class on Nov 10 ## Factoring Polynomials ### Example 1 To factor $x^2 + bx + c$, find two numbers $r$ and $s$ such that $r + s = b$ and $rs = c$. $x^2 + 5x + 6 = (x + 3)(x + 2)$ $x^2 - 5x + 6 = (x - 3)(x - 2)$ $x^2 + 5x - 6 = (x + 6)(x - 1)$ $x^2 - 5x - 6 = (x - 6)(x + 1)$ ### Example 2 Factor $2x^2 - x - 1$. We can multiply by 2 to get $4x^2 - 2x - 2$. Let $y = 2x$, then $y^2 - y - 2 = (y - 2)(y + 1) = (2x - 2)(2x + 1)$. Since we multiplied by 2 at the start, we divide by 2 to get $(x - 1)(2x + 1)$. We can also do this by trial and error. We know that the factors must be of the form $(2x \pm a)(x \pm b)$ ### Example 3 Difference of squares: $a^2 - b^2 = (a - b)(a + b)$ $4x^2 - 25 = (2x - 5)(2x + 5)$ ### Example 4 Difference of cubes; $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ Sum of cubes: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ ### Example 5 $x^5 - x = x(x^4 - 1) = x(x^2 - 1)(x^2 + 1) = x(x - 1)(x + 1)(x^2 + 1)$ ### Example 6 $(x + 1)^2 - 4 = ((x + 1) - 2)((x + 1) + 2) = (x - 1)(x + 3)$ ### Example 7 $x^4 + x^2 - 6 = (x^2 + 3)(x^2 - 2)$ ## Polynomial division ### Setup $\frac{x^3 - 3x^2 + 4x - 5}{x - 2}$ ### Solution * $x - 2 \>\>\> x^2$ * $x^3 - 3x^2 + 4x - 5$ * $\>\>\> x^3 - 2x^2$ * $\>\>\> -x^2 + 4x - 5$ * $x - 2 \>\>\> -x$ * $\>\>\> -x^2 + 2x$ * $\>\>\> 2x - 5$ * $x - 2 \>\>\> 2$ * $\>\>\> 2x - 4$ * $\>\>\> -1$ ### Answer $x^2 - x + 2 - \frac{1}{x - 2}$