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121xy + 11zx

p - p + 15pR^3

= 121xy + 11zx= 11x(11y) + 11x(z)= 11x(11y + z)

Answer: 11x(11y + z)-----------

= p - p + 15pR

= p() + p(- ) + p(15R)

= p( - + 15R)

Factorization of Quadratic Trinomials

TheDistributive Lawis used in reverse to factorise a quadratic trinomial, as illustrated below.

Consider the expansion of (x + 2)(x + 3).

We notice that:

5, the coefficient ofx, is the sum of 2 and 3.

6, the independent term, is the product of 2 and 3.
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Note:

The product of two linear factors yields aquadratic trinomial; and thefactorsof a quadratic

trinomial are linear factors.

Now consider the expansion of (x + a)(x + b).

Coefficient ofx = a + b = Sum ofa and b.Independent term = ab = Product ofa and b.

In general:

To factorize aquadratic trinomial, find two numbers whose sum is equal to thecoefficientofx,

and whose product is equal to theindependent term.

Example 12

Solution:
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Check:

Factors by Grouping 'Two and Two'

Now, consider the expression 7x + 14y + bx + 2by. Clearly, there is nofactorcommon to every

term. However, it is clear that 7 is acommon factorof the first two terms and b is a common

factor

of the last two terms. So, the expression can be grouped into two pairs of two terms as shown.
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This factorisation technique is called grouping 'Two and Two'; and it is used to factorise anexpression consisting of four terms.

Factorisation using the Common Factor

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We know that:

a(b + c) = ab + ac

The reverse process, ab + ac = a(b + c), is called taking out the common factor.

Consider the factorisation of the expression 5x + 15.

Note that thecommon factor5 has been taken out and placed in front of the brackets. Theexpression inside the brackets is obtained by dividing each term by 5.

In general:

To factorise an algebraic expression, take out thehighest common factorand place it in front of

the brackets. Then the expression inside the brackets is obtained by dividing each term by the

highest common factor.

Example 6

Solution:
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Note:

The process of taking out acommon factoris of great importance in algebra. With practice youwill be able to find thehighest common factor(HCF) readily and hence factorise the given

expression.

Example 7

Solution:

Note:

We can check the answer by using theDistributive Law.
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Highest Common Factor of Algebraic ExpressionsThehighest common factor (HCF)of algebraic expressions is obtained in the same way as that of

numbers.

Example 5

Solution:
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Example 8

Solution:

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Simplify3log2(x)4log2(x+ 3) + log2(y).

I will get rid of the multipliers by moving them inside as powers:

3log2(x)4log2(x + 3) + log2(y)= log2(x

3)log2((x + 3)

4) + log2(y)

Then I'll put the added terms together, and convert the addition to multiplication:

log2(x3

)log2((x + 3)4

) + log2(y)= log2(x3) + log2(y)log2((x + 3)4)= log2(x

3y)log2((x + 3)

4)

Then I'll account for the subtracted term by combining it inside with division:

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Let logb(2) = 0.3869, logb(3) = 0.6131, andlogb(5) = 0.8982. Using these values,

evaluate logb(10).

Since 10 = 2 5, then:

logb(10) = logb(2 5) = logb(2) + logb(5)

Since I have the values forlogb(2) and logb(5), I can evaluate:

logb(2) + logb(5) = 0.3869 + 0.8982 = 1.2851

Then logb(10) = 1.2851.

Let logb(2) = 0.3869, logb(3) = 0.6131, andlogb(5) = 0.8982. Using these values, evaluate

logb(9).

Since 9 = 32, then:

logb(9) = logb(32) = 2logb(3)

Since I have the value forlogb(3), then I can evaluate:

2logb(3) = 2(0.6131) = 1.2262

Then logb(9) = 1.2262.


logb(7.5).

This one is a bit more complicated, but, after fiddling with the numbers for a bit, I notice that 7.5 =

15 2, so:

logb(7.5) = logb(15 2) = logb(15)logb(2)

And 15 = 5 3, so: Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

logb(15)logb(2)= [logb(5) + logb(3)]logb(2)= logb(5) + logb(3)logb(2)

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And now I can evaluate:

logb(5) + logb(3)logb(2)= 0.8982 + 0.61310.3869= 1.1244

Then logb(7.5) = 1.1244.


logb(6).

Since 6 = 2 3, then:

logb(6) = logb(2 3) = logb(2) + logb(3)

Since I have these values, I can evaluate:

logb(2) + logb(3) = 0.3869 + 0.6131 = 1.0000

Then logb(6) = 1.0000.

Hmm... that was interesting. I got that logb(6) = 1. UsingThe Relationship, I get:

logb(6) = 1b

1= 6

b = 6

So now I know that their mysterious unnamed base "b" was actually 6! But they will not usually give youproblems that let you figure out the base like this.

Expand log4(16

/x).

I have division inside the log, which can be split apart as subtraction outside the log, so:

log4(16

/x ) = log4(16)log4(x)

The first term on the right-hand side of the above equation can be simplified to an exact value, byapplying the basic definition of what alogarithmis:

log4(16) = 2

Then the original expression expands fully as:

log4(16

/x ) = 2log4(x)

Always remember to take the time to check to see if any of the terms in your expansion (such as the

log4(16) above) can be simplified.
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Expand log5(x3). Copyright Elizabeth Stapel 2002-2011 All Rights Reserved

The exponent inside the log can be taken out front as a multiplier:

log5(x3) = 3 log5(x) = 3log5(x)

Expand the following:

The 5 is divided into the 8x4, so split the numerator and denominator by using subtraction:

Don't take the exponent out front yet; it is only on the x, not the 8, and you can only take the

exponent out front if it is "on" everything inside the log. The 8 is multiplied onto thex4, so split the

factors by using addition:

log2(8x4)log2(5) = log2(8) + log2(x

4)log2(5)

Thex has an exponent (which is now "on" everything inside its log), so move the exponent outfront as a multiplier:

log2(8) + log2(x4)log2(5)

= log2(8) + 4log2(x)log2(5)

Since 8 is a power of2, I can simplify the first log to an exact value:

log2(8) + 4log2(x)log2(5)

= 3 + 4log2(x)log2(5)

Each log contains only one thing, so this is fully simplified. The answer is:

3 + 4log2(x)log2(5)

Expand the following:

Use the log rules, and don't try to do too much in one step:

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