Factorization H.C.F L.C.M Simplification and Square roots
Factor
- A number that divides another number perfectly.
- A divisor that exactly divides by dividend then the divisor is known as a factor of the dividend.
$50÷2$ $=25$
Hence $2$ is the factor of $50$
The factor of the algebraic expressions
- An algebraic expression that divides exactly into another expression is known as the factor of the expression.
Prime number
- A natural number greater than $1$ that has only two factors $1$, and itself.
$2$, $3$, $5$, 7$$, $11$, $13$, $17$,....
Similarly in Algebra
- An algebraic expression that has only two factors one and itself is known as a prime factor.
$\left( x-a \right)$, $\left( x+a \right)$, $\left( x^2+a^2 \right)$
Composite Number
- A natural number greater than $1$ has more than two factors.
$4$, $6$, $8$, $9$, $10$, $12$, $14$, $15$, $16$, $18$, $20$...
Similarly in Algebra
$\left( x+a \right)^2$, $\left( x^2-a^2 \right)$, $\left( x+a \right)^3$, $\left( x-a \right)^2$
- A composite number or expression can be factorized into prime factors.
As
$20=$ $1\times 2$ $\times 2$ $\times 5$
$\left( x^2-a^2 \right)$ $=\left( x-a \right)$ $\left( x+a \right)$
Prime Factorization
- Prime factorization is a way of showing a composite number or expression as the product of a prime number or expression.
- The process of finding factors of a given expression is called factorization.
- Factorization means finding factors.
$1.$ Factors of the expression of the type
$ka$ $+kb$ $+kc$ $=k$ $\left( a+b+c \right)$
where $k$ is any constant number or any algebraic term
e.g.
$6x^2y^2$ $+12xy^3$ $-30x^3y$
$6xy$ $\left( xy+2y^2-5x^2 \right)$
$2.$ Factors of the expression of the type
$ac$ $+ad$ $+bc$ $+bd$
$a(c+d)$ $+b(c+d)$
$(c+d)$ $(a+b)$ $=(a+b)$ $(c+d)$
Method
Step $1:$ Regroup terms (such that terms with common factors are together)
Step $2:$ Write each term as a product of irreducible factors
Step $3:$ Apply distributive law
Step $4:$ Irreducible factors are obtained
Example: Resolve into factors
$i)$ $2x^2y^3$ $+2x^4y$ $-3x^3y^2$ $-3xy^4$
solution:
$=xy$ $(2xy^2$ $+2x^3$ $-3x^2y$ $-3y^3)$
$=xy$ $\{2x(y^2+x^2)-3y(x^2+y^2)\}$
$=xy$ $(x^2+y^2)(2x-3y)$
$3)$ Perfect square
Perfect squares are of the form
$a^2$ $+2ab$ $+b^2$ $=(a+b)^2$
$a^2$ $-2ab$ $+b^2$ $=(a-b)^2$
Identification and Solution
Identify and solve a trinomial is a perfect square
Step $1:$ Notice the first term is a square, take its square root.
Step $2:$ Notice the last term is a square, take its square root.
Step $3:$ Multiply the results of the first 2 steps and double that product, if the result is the middle term of the trinomial, then the expression is a perfect square.
Step $4:$ The binomial in the solution is the sum or difference of the
square roots calculated in steps $1$ and $2$.
The middle term of the trinomial has the same sign as the sign between the
components of the binomial.
Example: Resolve into factors
$i)$ $9$ $-6(a-3b)$ $+(a-3b)^2$ $+(a-3b)^4$
solution:
Take the square root of the first term and last term
$\sqrt{9}$ $=\pm 3$ and $\sqrt {(a-3b)^4}$ $=\pm (a-3b)$
Multiplying the result and doubling that
$2$ $\times 3$ $(-(a-3b))$ $=-6(a-3b)$
$\left( 3-\left( a-3b \right)^2 \right)^2$ or $\left( \left( a-3b \right)^2-3 \right)^2$
$ii)$ $a^2b^2$ $-6ab$ $+9$
Solution:
Take the square root of the first term and last term
i.e $\sqrt {a^2b^2}$ $=\pm ab$ and $\sqrt {9}$ $=\pm 3$
Multiplying the result and doubling that
$2(ab)(-3)$ $=-6ab$
$(ab)^2$ $+2(ab)(-3)$ $+(-3)^2$
$(ab-3)^2$
$4.$ Difference of squares
The difference between squares is the form
$$a^2-b^2 =(a+b)(a-b)$$
Because there is no something like the middle word, they are more simpler to identify than the perfect squares
Notice why is no middle term
$(a+b)(a-b) =a^2-ab+ab-b^2$ $=a^2-b^2$
Two middle terms can be canceled
Answer "yes" to four questions if an expression is a difference of squares.
Method
$1.$ Are there only two terms? "Yes"
$2.$ Do they have a $"-"$ symbol between them?
$3.$ Is the first term a square? if so, take its square root.
$4.$ Is the second term a square?
Examples:
$i)$ $\left( ax^4-\frac{a}{16} \right)$
Solution:
$=a$ $\left( \left( ax^2 \right)^2$ $-\left( \frac{1}{4}
\right)^2 \right)$
$=a$ $\left( x^2+\frac{1}{4} \right)$ $\left( x^2-\frac{1}{4} \right)$
$ii)$ $\left( a^4b^4-144c^2 \right)$
Solution:
$=a$ $\left( \left( a^2b^3 \right)^2$ $-\left( 12c \right)^2 \right)$
$=\left( a^2b^3+12c \right)$ $\left( a^2b^3-12c \right)$
$iii)$ $\left( a-b \right)^2$ $-9c^2$
Solution:
$=\left( a-b \right)^2$ $-\left( 3c \right)^2$
$=\left( a-b+3c \right)$ $\left( a-b-3c \right)$
5. Factorizing Trinomial
- The factorization of a trinomial that is neither a perfect square nor a difference of squares is a frequent issue in elementary algebra.
$x^2$ $+(p+q)x$ $+(pq)$ $=(x+p)(x+q)$
- The sign after $x^2$ is known as sign $1$ and the sign after $x$ is sign $2$
$x^2$ $+ax$ $+b$ $=x^2$ $+(p+q)x$ $+(pq)$ $=(x+p)(x+q)$, where $a$ $=(p+q)$ and $b$ $=pq$
Simple Case Method
Let $x^2$ $+15x$ $-100$
Step $1:$ Set up parenthesis for a pair of binomials. Put $x$ in the left-hand position of each binomial.
e.g. $=$ $(x$ $)(x$ $)$
Step 2: Put the sign $1$ in the middle position in the left binomial.
$=$ $(x+$ $)(x$ $)$
Step 3: Multiply sign $1$ and sign $2$ to get the sign for the right binomial.
$=$ $(x+$ $)(x$ $)$
$∴$ $+\times -$ $=-$
$=$ $(x+$ $)(x-$ $)$
Remember that:
$+$ $\times +$ $=+$, $-$ $\times -$ $=+$
$-$ $\times +$ $=-$, $+$ $\times -$ $=-$
Step $4:$ Find the L.C.M of constant number and find two numbers such that
$(a)$ Multiply to get the last term (constant)
$100$ $=1\times 2$ $\times 2$ $\times 5$ $\times 5$
$(20)(-5)$ $=-100$
$(b)$ Add to get middle term (coefficient of $x$).
$20-5$ $=15$
$(x+20)(x-5)$
$2nd$ Method
$r^6$ $-10r^3$ $+16$
Step $1:$ Find the L.C.M of the last term(constant) and write as all prime numbers
$16$ $=1\times 2$ $\times 2$ $\times 2$ $\times 2$
Step 2: Find two numbers such that
$(a)$ Multiply to get the last term(constant).
$(-2)(-8)$ $=16$
$(b)$ Add to get the middle term(coefficient of $x$)
$-2-8$ $=-10$
Step $3:$ Spilt the middle term into two terms with coefficients equal to the values found in step $2$
$r^6$ $-2r^3$ $-8r^3$ $+16$
Step 4: Group the terms into pairs
$(r^6-2r^3)$ $-(8r^3-16)$
$r^3(r^3-2)$ $-8(r^3-2)$
$(r^3-2)(r^3-8)$
Example:
$(i)$ $ax^4$ $-20ax^2y^2$ $-96ay^4$
Solution:
$96$ $=1\times 2$ $\times 2$ $\times 2$ $\times 2$ $\times 2$ $\times 3$
$-96$ $=ab$ $20$ $=a+b$
$(2$ $\times 2$ $\times 2$ $\times 3)$ $\times (2$ $\times 2)$ $=96,$ $a$ $=-24$, and $b$ $=4$
$-24$ $+4$ $=20$
$=ax^4$ $-24ax^2y^2$ $+4ax^2y^2$ $-96ay^4$
$=ax^2(x^2$ $-24y^2)$ $+4ay^2(x^2y$ $-24y^2)$
$=(x^2$ $-24y^2)$ $(ax^2$ $+4ay^2)$
$=a(x^2$ $-24y^2)$ $(x^2$ $+4y^2)$
$(ii)$ $(a+b)^2$ $+20(a+b)$ $+36$
Solution:
$36$ $=1$ $\times 2$ $\times 2$ $\times 3$ $\times 3$
$2$ $\times 18$ $=36$
$2+18$ $=20$
$=(a+b)$ $+2(a+b)$ $+18(a+b)$ $+36$
$=(a+b)$ $\left\{ \left( a+b \right)+2 \right\}$ $+18$ $\left\{ (a+b)+2 \right\}$
$=(a+b+2)$ $(a+b+18)$
$6.$ Expression of the type
$(a)$ $a^2$ $+2ab$ $+b^2$ $-c^2$
$(b)$ $a^2$ $-2ab$ $+b^2$ $-c^2$
$(a)$ $a^2$ $+2ab$ $+b^2$ $-c^2$
$(b)$ $a^2$ $-2ab$ $+b^2$ $-c^2$
$(a+b)^2$ $-c^2$ $=(a-b+c)$ $(a-b-c)$
Identification
$(1)$ Are they have four terms?
$(2)$ Are three terms squared?
$(3)$ Is one or two terms negative?
$(4)$ Is one term double of $1^{st}$ term and $2^{nd}$ term?
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