Factoring Algebraic Expressions

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Factoring Algebraic Expressions

Polynomials and Factoring

I will ask you three algebraic expression to be factorized. It will be useful to review binomial, trinomial, quadratic and cuing relationships such as :
(x+y)(x-y) = …
(x +y)^2 = …
(x - y)^2 = …
(x+ y)^3 = …
x^3 + y^3 = …, etc
Now, factorize the following 3 problems, showing the methods used, for the benefit of other students who need to learn factoring procedures.

1) 20 Z^2 - 53 ZX + 18 X^2

2) 216 P^3 + 125 Q^3

3) 6X^2 + 7X = 3

Posted 170 day ago

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	Best Answer I like math_fanatic’s Answers. It is obvious that she has a good algebraic background and wants to learn more. I will show details for question # 1 which will be somewhat easy to understand. The product of (a+b)(c+d) is ac +ad + bc + bd, these 4 terms are known as F O I L in basic algebra. F, represents the product of First characters “a” and “c” in the two factors, O, represents the product of outside characters, “a” and “d” I, represents the product of inside characters, “b” and “c”, and L, represents the product of last characters in the two factors, “b” and “d”. Now, let us look at question # 1, i.e. 20Z^2 – 53 ZX + 18 X^2, We need to split 53ZX in two parts, 8 and 45 (if you carefully follow this solution, you will see why 8 and 45), and then use F O I L to factor the given expression. We write our problem as: 20Z^2 – 45 ZX – 8ZX + 18X^2, and we form two factors as shown below: ( Z )( Z ). The best guess to split 20 is 4 and 5 and 18 as 9 and 2. However, if we find that 4 and 5 or 9 and 2 do not work, we can split 20 into 2 and 10 and 18 into 6 and 3 and try again. Adding 4 and 5 to our two factors we get: ( 4Z )( 5Z ). We see that O and I terns are both negative, therefore we get: ( 4Z – 9X )( 5Z – 2X), multiplying we get: 20 Z^2 – 8ZX – 45XZ + 18X^2 or 20 Z^2 – 53ZX + 18X^2, which is what we need. That is all! If you liked this lecture, just let me know, and I will detail the other two questions. Solved by: Professor rajbaba78. Posted 170 day ago ( 0 ) ( 0 ) Report Abuse

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Answers (5)

matherapist

I find your explanation confusing. You are basically using a method called factoring by grouping.
First -multiply the coeficients of the first and last terms 20 x 18 = 360

Second- write all 2 number combinations which when multiplied together = 360. ie
(360,1), (180,2)......... (60,6), (45,8), ......(15,24)..

Third- Because the product was +360 we know that both factors are either positive or negative and must sum to -53 (coeficient of the middle term.

We can then write the equation as you specify
20Z^2 - 45ZX - 8ZX +18X^2

We can then apply the distributive law to 20Z^2-45ZX = 5Z(4Z-9X)
Similarly for the last 2 terms we get -2X(4Z-9X). Combining them we have
5Z(4Z-9X)-2X(4Z-9X) = (4Z-9X)(5Z-2X)

Very little trial and error except for step 2.

There is a method which I use that involves the Foil concept but the format of this forum doesn't allow a good explanation and is usually very fast if you are good at mental arithmetic with simple numbers.

Basically you write all the 2 number multipliers of the First coeficient and last coeficient.
Sort of cross multiply and use the sign info from the original problem and find the inner and outer products whch when combined equals the coeficient of the second term.

Posted 169 day ago

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I like math_fanatic’s Answers. It is obvious that she has a good algebraic background and wants to learn more. I will show details for question # 1 which will be somewhat easy to understand.

The product of (a+b)(c+d) is ac +ad + bc + bd, these 4 terms are known as F O I L in basic algebra.

F, represents the product of First characters “a” and “c” in the two factors,
O, represents the product of outside characters, “a” and “d”
I, represents the product of inside characters, “b” and “c”, and
L, represents the product of last characters in the two factors, “b” and “d”.
Now, let us look at question # 1, i.e. 20Z^2 – 53 ZX + 18 X^2, We need to split 53ZX in two parts, 8 and 45 (if you carefully follow this solution, you will see why 8 and 45), and then use F O I L to factor the given expression. We write our problem as: 20Z^2 – 45 ZX – 8ZX + 18X^2, and we form two factors as shown below:
( Z )( Z ). The best guess to split 20 is 4 and 5 and 18 as 9 and 2. However, if we find that 4 and 5 or 9 and 2 do not work, we can split 20 into 2 and 10 and 18 into 6 and 3 and try again. Adding 4 and 5 to our two factors we get:
( 4Z )( 5Z ). We see that O and I terns are both negative, therefore we get:
( 4Z – 9X )( 5Z – 2X), multiplying we get: 20 Z^2 – 8ZX – 45XZ + 18X^2 or
20 Z^2 – 53ZX + 18X^2, which is what we need.
That is all!
If you liked this lecture, just let me know, and I will detail the other two questions.
Solved by: Professor rajbaba78.

Posted 170 day ago

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math_fanatic

Students aren't going to answer your questions, especially hard ones like this.
The only people who are answering these kinds of questions are the tutors.
If you like, we can ignore you and then no one will answer your questions.

If you have a genuine question that YOU need help with then ask it.
Otherwise, you're just providing brain teasers for the tutors.

AND you're asking us to show the procedures for this type of problem using the
limited symbols available on a qwerty keyboard to show students how to do it.

Here are my answers:
1) (4z - 9x) (5z - 2x) I used trial and error to come up with this answer. I cannot explain
my algorithm in this forum. If you have a simple algorithm for solving this kind of problem
that you can post in this forum, by all means do so and enlighten the rest of us. Just post it
in another thread because we don't get alerts when new answers are added to questions
we've answered.

2) This is the sum of two cubes which will take on the form (a + b) (a^2 - ab + b^2)
In this case a = 6p (the cube root of 215p^3) and b = 5q (the cube root of 125q^3)
Using the sum of cubes formula, the answer is (6p + 5q) (36p^2 - 30pq + 25q^2)

3) 6x^2 + 7x = 3
subtract 3 from both sides to get a quadratic equation: 6x^2 + 7x - 3 = 0
The factored form is: (2x + 3) (3x -1) = 0 [Again I used trial and error]
Thus x = 1/3 or -3/2

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matherapist

I sort of agree with Dr Bob on this. You claim to be a math professor complete with the degrees. My explanation of factoring would be too complex for a student to read without doing easier problems, but your answers are :

1 (4Z-9X)(5Z-2X) < 45 sec
2 (6P+5Q)(36P^2-30PQ+25Q^2) < 40 sec
3) (3X-1)(2X-3) < 25sec

If you want to teach factoring to students, start with simpler problems.

I do not use the methods described in typical texts, but rather one I learned in 1952, my freshman year in HS. The cubic is the stnd formula
a^3 + b^3= and substituting 6P and 5Q for a and b

The other answers can be obtained by using the Quadratic formula or my method.

So for exercise try factoring
x^5+x^2y^3-x^3y2-y^5=0

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Doctor Bob

Do your own work, good buddy, and do your own reviewing.
What have you tried so far?
What do you understand?
What don't you understand.
I don't see any effort on your part to do any of your own work i.n this question or on any previous questions
I will not be a party to cheating.

Posted 170 day ago

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