G8_Factoring

Factorization

Why do we need factoring?
The area of a rectangular photo frame \(\text{🖼️}\) is given as
\(x^2 + 15x + 54\).
To find the possible dimensions of this photo frame, we need to perform factoring.

🖼️

Factors

The product of two algebraic expressions are given below.

1. \((x) (x + 2) = x^2 + 2x\)
2. \((x + 2)(x + 3) = x^2 + 5x + 6\)
3. \((x + 2)(x + 1) = x^2 + 3x + 2\)

Study them to answer following questions.

1. Do the expressions \(x\) and \((x + 2)\) divide \(x^2 + 2x\) exactly? If so, what is called the relation of \(x\) and \((x + 2)\) to \((x^2 + 2x)\)?
2. Do the expressions \((x + 2)\) and \((x + 3)\) divide \(x^2 + 5x + 6\) exactly? If so, what is called the relation of \((x + 2)\) and \((x + 3)\) to \((x^2 + 5x + 6)\)?
3. Do the expressions \((x + 2)\) and \((x + 1)\) divide \(x^2 + 3x + 2\) exactly? If so, what is called the relation of \((x + 2)\) and \((x + 1)\) to \((x^2 + 3x + 2)\)?

Things you need to know?

The algebraic expressions which divide the given algebraic expression exactly are called factors of the given expression.
If an algebraic expression is expressed as the product of its factors, then it is called factorization.

After factoring a polynomial, if any of its factors divides the original polynomial, the remainder will be zero.

What are the common methods of factorization?
The common methods of factorization are as follows

1. Factoring monomials
2. Factoring common element
3. Factoring by grouping
4. Factoring by identities
5. Factoring by FOIL

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Factoring monomials

Factoring a monomial expression means breaking the given term into its simpler multiplicative components. This involves expressing both the numerical part and the variables in terms of their smaller factors.

Factorize \(20x^2\)

When factoring \(20x^2\)
the number \(20\) is written as the product of its prime factor
\(2, 2, 5\)
and \(x^2\) is written as
\(x \cdot x\)
So, the factorization of \(20x^2\) is
\(20x^2 = 2 \cdot 2 \cdot 5 \cdot x \cdot x\)

|
| Expression --------->  20x2
|                        / \
|                       /   \
|                      /     \
| Factors --------->  20      x2
|                     /|\     / \
|                    / | \   /   \
|                   /  |  \ /     \
| Factors -------> 2   2  5 x     x
|

Factoring common element

The process of factoring using a common element is called factoring with common element. In this method, a common factor that appears in all terms of a polynomial expression is taken out (factored out). For example
Expanding the expression \(3(x + 1)\) gives: \(3x + 3\).
Now
Factoring \(3x + 3\) gives: \(3x + 3 = 3(x + 1)\)

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Factoring by grouping

The method of factoring a polynomial by organizing it into two or more groups and then taking out the common factor from each group is called factoring by grouping. This method is especially useful for polynomial expressions with four terms. For example, to factor \(ax + ay + bx + by\).

1. organizing the terms into two groups
   \((ax + ay) + (bx + by)\)
2. Take out the common factor from each group
   \(a(x + y) + b(x + y)\)
3. Here, \((x + y)\) is common in both terms, so factor it out
   \((x + y)(a + b)\)

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FOIL method

Factoring \(ax^2+bx+c\) by FOIL method

In school mathematics, FOIL is an acronym for the standard method of multiplying two binomials. The word FOIL represents four terms of the product:

1. First ("first" terms of each binomial are multiplied together)
2. Outer ("outside" terms are multiplied)
3. Inner ("inside" terms are multiplied)
4. Last ("last" terms of each binomial are multiplied)

The general form is
\((a+b)(c+d)= ac+(ad+bc)+bd\) (First+outside+Inside+Last)(i)
Now, given the quadratic equation
\(Ax^2+Bx+C\) (ii)
If we set (i) and (ii) in order, we get
\(A=ac, B=ad+bc, and C=bd\)
Given
\(B=ad+bc= \alpha + \beta\), say
Then
\(Ax^2+Bx+C\)
or\(Ax^2+( \alpha + \beta)x+C\)
where
\(AC= ad \cdot bc= ad \cdot bc= \alpha \beta\)

Example: Visualization of algebraic factors

1. \(x^2+5x+6\)
   One piece of \(x^2\) square units, five of \(x\) square units and six of \(1\) square units are arranged.
   The answer is : \(x^2+5x+6=(x+2)(x+3)\)
2. \(x^2-5x+6\)
   One piece of \(x^2\) square units, five of \(-x\) square units and six of \(1\) square units are arranged.
   The answer is : \(x^2-5x+6=(x-2)(x-3)\)
3. \(x^2+5x-6\)
   One piece of \(x^2\) square units, five of \(x\) square units and six of \(-1\) square units are arranged.
   The answer is : \(x^2+5x-6=(x-1)(x+6)\)
4. \(x^2-5x-6\)
   One piece of \(x^2\) square units, five of \(-x\) square units and six of \(-1\) square units are arranged.
   The answer is : \(x^2-5x-6=(x+1)(x-6)\)

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Factoring by identities

The method of factoring a polynomial using identities or formulas is called factoring by identities or formula. In this method, the given polynomial is compared with standard algebraic identities, and the factoring is done by applying those formulas.

\(a^2-b^2 = (a-b)(a + b)\)

The expression
\(a^2-b^2 = (a-b)(a + b)\)
can be visualized using algebra tiles. This is a difference of squares, and its factorization results in \((a-b)(a+b)\).

1. Take one large square tile to represent \(a^2\).
   This tile has both length and width equal to \(a\), so its area is \(a \times a = a^2\).
2. Cut out one small square tile to represent \(b^2\).
   This tile has both length and width equal to \(b\), so its area is \(b \times b = b^2\).
3. Subtract the small square \(b^2\) from the large square \(a^2\).
   This represents the expression \(a^2 - b^2\).
4. Rearranging the remaining area, to form a rectangle, in which
   length is \((a + b)\)
   width is \((a - b)\)

For example
\(x^2 - 9 = (x - 3)(x + 3)\)

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\((a + b)^2=a^2 + 2ab + b^2 \)

The expression \((a + b)^2\) can be directly visualized using algebra tiles. This expression is the sum of \(a^2, \,ab\), and \(b^2\), and its factorization is \((a + b)^2\).
To illustrate this using algebra tiles, follow these steps.

1. Take one large square tile to represent \((a+b)^2\).
   This tile has both length and width equal to \(a+b\), so its area is
   \((a+b) \times (a+b) = (a+b)^2\)
2. separate the dimension for \(a\) and \(b\)
   This will create four tiles, one of \(a^2\), one of \(b^2\) and rest two of \(ab\) as area.
3. Rearranging the tiles, to form
   \(a^2 + 2ab + b^2 = (a + b)^2\)

For example
\(x^2 + 6x + 9 = (x + 3)^2\)

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Algebraic tile of\((a-b)^2=a^2-2ab+b^2\)

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\((a+b)^3 =a^2+3a^2b+3ab^2+b^3)\)

बीजगणितीय टायल (algebra tiles) द्वारा \((a + b)^3\) लाई प्रत्यक्ष रूपमा बुझाउन, घन (cube) को प्रयोग गरिन्छ। \((a + b)^3\) लाई बीजगणितीय टायल प्रयोग गरेर देखाउदा,

1. \((a + b)^3\): सबै टायलहरूले मिलेर एउटा घन (cube) बनाएको छ जसको लम्बाई चौडाई र उचाई सबै \((a+b)\) छ ।
2. सबैभन्दा ठूलो घनले \( a^3 \) लाई प्रतिनिधित्व गर्छ, र यसलाई तल्लो-बायाँ कुनामा राखिएको छ।
3. तिन वटा \( a^2b \) आयतन भएका षडमुखाहरु छन, जुन क्रमश: एउटा दाँया, एउटा माथी र एउटा पछाडी, प्रयोग भएको छ ।
4. तिन वटा \( ab^2 \) आयतन भएका षडमुखाहरु छन, जुन क्रमश: एउटा दाँया, एउटा माथी र एउटा अगाडी, प्रयोग भएको छ ।
5. सबैभन्दा सानो घन \( b^3 \) लाई प्रतिनिधित्व गर्छ र यसलाई माथिल्लो-दायाँ कुनामा राखिएको छ।

\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

The large cube has a volume of \( a^3 \). It is made up of four smaller parts (a cube and three cuboid. So, we can write an expression for the volume of \((a^3-b^3)\), as
\( a^3-b^3=V_2+ V_3+ V_4 \)
Here
\(V_1\) volume= \(b^3\)
\(V_2\) volume= \((a-b)b \cdot b = (a-b)b^2 \)
\(V_3\) volume= \((a-b)b \cdot a = (a-b)ab\)
\(V_4\) volume= \((a-b)a \cdot a = (a-b)a^2\)
Therefore
\(a^3 - b^3=V_2+ V_3+ V_4 \)
or\(a^3 - b^3=(a-b)a^2+(a-b)ab+(a-b)b^2 \)
or\(a^3 - b^3=(a-b)(a^2+ab+b^2) \)

उदाहरण
\(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)

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For Q.No.6 (a) in BLE Examination

1. \( (ax + by)^2 \) लाई विस्तारित रूपमा लेख्नुहोस्।
   Write down \( (ax + by)^2 \) in expanded form. [1U]
The expanded form is
\( (ax + by)^2 \)\((a+b)^2=a^2+2ab+b^2\)

or\( (ax )^2+2(ax)(by)+(by)^2 \)\(a=(ax), b=(by)\)

or\( a^2x^2 + 2abxy + b^2y^2 \)
2. \( (x + 2)^3 \) लाई विस्तारित रूपमा लेख्नुहोस्।
   Write \( (x + 2)^3 \) in expanded form. [1U]
The expanded form is
\( (x + 2)^3 \)\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

or\( x^3 + 3x^2(2) + 3x(2)^2 + (2)^3 \)\(a = x, b = 2\)

or\( x^3 + 6x^2 + 12x + 8 \)
3. \( (x^2 - 1) \) लाई खण्डित रूपमा लेख्नुहोस्।
   Write \( (x^2 - 1) \) in factorized form. [1U]
The factorized form is
\( x^2 - 1 \)\( a^2 - b^2 = (a - b)(a + b) \)

or\( (x - 1)(x + 1) \)\( a = x, b = 1 \)
4. \( P^2 - \dots + 36 \) लाई पूर्ण वर्ग बनाउन खाली ठाउँमा कुन पद राख्नुपर्छ?
   What term should be inserted in \( P^2 - \dots + 36 \) to make it a perfect square? [1A]
The term to be inserted is
\( P^2 - \dots + 36 \)Perfect square: \( (a - b)^2 = a^2 - 2ab + b^2 \)

or\( P^2 - \textcolor{red}{2(a)(b)}+6^2 \)
or\( P^2 - \textcolor{red}{2(P)(6)}+6^2 \)
or\( P^2 - \textcolor{red}{12P}+6^2\)
The term that should be inserted in \( P^2 - \dots + 36 \) to make it a perfect square is \(\textcolor{red}{12P}\)
5. \( 4a^2 + \dots + y^2 \) लाई पूर्ण वर्ग बनाउन खाली ठाउँमा कुन पद राख्नुपर्छ?
   What term should be inserted in \( 4a^2 + \dots + y^2 \) to make it a perfect square? [1A]
The term to be inserted is
\( 4a^2 + \dots + y^2 \)Perfect square: \( (a + b)^2 = a^2 + 2ab + b^2 \)

or\( (2a)^2 + \textcolor{red}{2(2a)(y)} + y^2 \)\( a = 2a, b = y \)

or\( 4a^2 + \textcolor{red}{4ay} + y^2 \)

The term that should be inserted in \( 4a^2 + \dots + y^2 \) to make it a perfect square is \(\textcolor{red}{4ay}\).
6. खण्डीकरण गर्नुहोस् (Factorize): \( x^2 + 4x \) [1U]
The factorized form is
\( x^2 + 4x \)\( a^2 + 2ab = a(a + 2b) \)

or\( x(x + 4) \)\( a = x, b = 2 \)
7. खण्डीकरण गर्नुहोस् (Resolve into factors): \( x^2 - \dfrac{1}{64y^2} \) [1U]
The factorized form is
\( x^2 - \dfrac{1}{64y^2} \)

or \( (x)^2 - \left ( \dfrac{1}{8y} \right )^2\) \( a^2 - b^2 = (a - b)(a + b) \)

or\( \left( x - \dfrac{1}{8y} \right)\left( x + \dfrac{1}{8y} \right) \)\( a = x, b = \dfrac{1}{8y} \)
8. खण्डीकरण गर्नुहोस् (Factorize): \( 4x^2 - 25z^2 \) [1U]
The factorized form is
\( 4x^2 - 25z^2 \)
or\( (2x)^2 - (5z)^2 \) \( a^2 - b^2 = (a - b)(a + b) \)

or\( (2x - 5z)(2x + 5z) \)\( a = 2x, b = 5z \)
9. खण्डीकरण गर्नुहोस् (Factorize): \( 9y^2 - 4 \) [1U]
The factorized form is
\( 9y^2 - 4 \)
or \( (3y)^2 - (2)^2 \) \( a^2 - b^2 = (a - b)(a + b) \)

or\( (3y - 2)(3y + 2) \)\( a = 3y, b = 2 \)
10. खण्डीकरण गर्नुहोस् (Factorize): \( 16 - x^4 \) [1U]
The factorized form is
\( 16 - x^4 \)
or \( (4)^2 - (x^2)^2 \) \( a^2 - b^2 = (a - b)(a + b) \)

or\( (4 - x^2)(4 + x^2) \)\( a = 4, b = x^2 \)

or\( (2 - x)(2 + x)(4 + x^2) \)\( 4 - x^2 = (2 - x)(2 + x) \)
11. खण्डीकरण गर्नुहोस् (Factorize): \( y^2 - xy + x - y \) [1U]
The factorized form is
\( y^2 - xy + x - y \)Group terms to factor

or\( (y^2 - xy) + (x - y) \)Group \( y^2 - xy \) and \( x - y \)

or\( y(y - x) - (y - x) \)Factor out \( y \) and \(-1\)

or\( (y - x)(y - 1) \)Factor out \( (y - x) \)
12. दिइएको बीजीय अभिव्यञ्जकहरूमा साझा गुणनखण्ड पत्ता लगाउनुहोस्: \( m(x - y) \), \( n(y - x) \)
    Find a common factor in the given algebraic expressions: \( m(x - y) \), \( n(y - x) \) [1K]
The common factor is
\( m(x - y) \), \( n(y - x) \)
or\( m(x - y) \) and \( -n(x - y) \)

The common factor is \( x - y \)
13. x को मान कति हुँदा तल दिइएको बीजीय भिन्नको मान अपरिभाषित हुन्छ ?
    For what value of x is the given algebraic fraction undefined? \( \dfrac{4}{x^2 - 9} \) [1HA]
For \(x=3\), the given algebraic fraction is undefined.