# Week 5 – Precalc 11

This week in Precalculus 11, we started the Solving Quadratic Equations Unit. We first reviewed factoring polynomial expressions.

Factoring: separating an expression into its components

Polynomial expression: an expression of numbers and variables being added, subtracted or multiplied (example: 2x)

Greatest common factor: the greatest number that can divide into all the terms in the expression (example: 2, 4, 6, 8, 10      GCF = 2)

Term: a number or a group of numbers being multiplied (example: 2$x^2$)

Binomial: a polynomial expression with 2 terms (example: x + 1)

Difference of squares: a polynomial expression in which subtraction takes place between 2 perfect square terms (example: 4x – 1)

Trinomial: a polynomial expression with 3 terms (example: $x^2$ + x + 1)

Conjugates: 2 terms with opposite addition/subtraction signs (example: 1 + 1 & 1 – 1)

To factor a polynomial expression, we first look for the greatest common factor between the terms and divide each term by that number.

Example:

2x + 2

= 2(x + 1)

If the expression is a binomial, we check if it is a difference of squares. When factored, a difference of squares results in conjugates.

Example:

4x – 1

= (2x + 1)(2x – 1)

If the expression is a trinomial, we check if it is in the form a$x^2$ + bx + c. If a = 1, we separate bx into 2 terms that multiply to c and add to bx.

Example:

$x^2$ + 2x + 1

= (x + 1)(x + 1)

= $(x + 1)^2$

If a ≠ 1, we separate bx into 2 terms that multiply to ac and add to bx, then find the greatest common factor of each side, and divide each term by that number.

Example:

2$x^2$ + 4x + 2

= 2$x^2$ + 2x + 2x + 2

= 2x(x + 1) + 2(x + 1)

= (2x + 2)(x + 1)

If the expression is in a different form than a$x^2$ + bx + c, the expression can sometimes be changed to this form.

Example:

$(x + 1)^2$ + 2(x + 1) + 1

= $a^2$ 2a + 1

= (a + 1)(a + 1)

= $(a + 1)^2$

= $[(x + 1) + 1]^2$

= $(x + 2)^2$

# Week 4 – Precalc 11

Radical expression: the term a$\sqrt[b]{c}$

Radical: a root sign (represented by √)

Coefficient: the number outside the radical, it is being multiplied by the radical

When adding or subtracting radical expressions, the radicals (the radicand and the root) must be like terms. Some radicals can be simplified to achieve like terms. The coefficients are then added or subtracted, and the radical remains the same.

Example:

$\sqrt{2}$ + $\sqrt{8}$

= $\sqrt{2}$ + $\sqrt{4}$$\sqrt{2}$

= $\sqrt{2}$ + 2$\sqrt{2}$

= 3$\sqrt{2}$

# Week 3 – Precalc 11

This week in Precalculus 11, we started the Absolute Value and Radicals unit. We first learned about the absolute value of a real number.

Absolute value of a real number: its distance from zero on a number line/the principal square root of its square

Principal square root: the positive (+) square root

Example: $\sqrt{1}$ = ±1     +1 = principal square root     -1 = negative (-) square root

The symbols “||” represent absolute value.

Example: <-0-1-2-3-4-5->     |0| = 0

|0|

= $\sqrt{0^2}$

= 0

This is a new concept I had not yet learned about and can add to my knowledge of Pre-Calculus.

# Week 2 – Precalc 11

This week in Precalculus 11, we learned about geometric sequences.

Geometric sequence: a list of terms in which each term is multiplied by a common ratio to equal the next term

Term: a number (represented by “$t_n$“, “n” being the term’s place in the geometric sequence)

Common ratio: a common number multiplied by each term to equal the next term (represented by “r”)

Example: 1, 2, 4, 8, 16     r = 2

To find the value of a term in a geometric sequence, we use the formula $t_n$ = $t_1$ · $r^{n - 1}$.

Example: Find $t_{10}$: 1, 2, 4, 8, 16

$t_{10}$ = ?     $t_1$ = 1     r = 2     n = 10

$t_n$ = $t_1$ · $r^{n - 1}$

($t_{10}$) = (1)$(2)^{(10) - 1}$

$t_{10}$ = 1 · $2^9$

$t_{10}$ = 1 · 512

$t_{10}$ = 512

This method of finding the value of terms in an geometric sequence is faster than extending the list.

# Week 1 – My Arithmetic Sequence

13, 26, 39, 52, 65…

$t_n$ = $t_1$ + d(n – 1)

($t_{50}$) = (13) + (13)[(50) – 1]

$t_{50}$ = 13 + 13 · 49

$t_{50}$ = 13 + 637

$t_{50}$ = 650

$S_n$ = $\frac{n}{2}$($t_1$ + $t_n$)

($S_{50}$) = $\frac{(50)}{2}$[(13) + (650)]

$S_{50}$ = 25 · 663

$S_{50}$ = 16575

# Week 1 – Precalc 11

This week in Pre-Calculus 11, we started the Sequences and Series unit. We first learned about arithmetic sequences.

Arithmetic sequence: a list of terms in which a common difference is added to each term to equal the next term

Term: a number (represented by “$t_n$“, “n” being the terms place in the arithmetic sequence)

Common difference: a common number added to each term to equal the next term (represented by “d”)

Example: 0, 1, 2, 3, 4     d = +1

t₁  t₂  t₃  t₄  t₅

To find the value of a term in an arithmetic sequence, we use the formula $t_n$ = t₁ + d(n – 1).

Example: Find t₁₀: 0, 1, 2, 3, 4

$t_{10}$ = ?     $t_1$ = 0     d = +1     n = 10

$t_n$ = t₁ + d(n – 1)

(t₁₀) = (0) + (1)[(10) – 1]

(t₁₀) = 0 + 1 · 9

(t₁₀) = 0 + 9

(t₁₀) = 9

This method of finding the values of terms in an arithmetic sequence is faster than extending the list.