Questions on Algebra: Polynomials, rational expressions and equations answered by real tutors!

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I think you are missing an exponent on the 2nd equation.
Cheers,
Stan H.

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Yu did not state directions but I assume you want to factor, right?
2ab + 2ay + bx + xy
When you have 4 terms, it is best to factor by groups.
2ab + 2ay = group 1
bx + xy = group 2
Factor each group individually.
2ab + 2ay = 2a(b + y)
===============================
bx + xy = x(b + y)
We now have this:
2a(b + y)x(b + y)
Notice that we have two quantities that are the same, which is (b + y).
We only need one of them.
Your final answer is the following two factors:
(2a + x)(b + y)

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(2x^2 all over 3 times 5 all over x)
= [(2x^2)/3] * [5/x]
Multiply the numerators and multiply the denominators.
= [10x^2/3x]
=====================
Cheers,
Stan H.

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L=3W
LW=108
3W*W=108
3W^2=108
W^2=108/3
W^2=36
W=SQRT36
W=6 IN. FOR THE WIDTH.
L=3*6=18 IN.FOR THE LENGTH.

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The formula for finding the area of a triangle is:
AREA = 1/2 times base times height
In short, we write it like this: A = 1/2 (b) (h)
The base is given to be 2x + 1.
The height is given to be x.
We are looking for A(x), which is read "A of x."
NOTE: A(x) does not mean A TIMES X. A lot of students make this mistake when it comes to functions.
We now have this function:
A(x) = 1/2 (2x + 1) ( x)
We are told to find A(5). This means to replace every x you see with 5 and simplify.
A(5) = 1/2 (2(5) + 1) (5)
A(5) = 55/2

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1. 16-m^2
(4+m)(4-m)
2. 9a^2b^2-100
(3ab+10)(3ab-10)
3. a^2-36+169
Using the quadratic equation we get
a=(36+-sqrt[-36^2-4*1*169])/281
a=(36+-sqrt[1,296-676])/2
a=(36+-sqrt620)/2
a=(36+-24.9)/2
a=(36+24.9)/2
a=60.9/2
a=30.45 ans.
a=(36-24.9)/2
a=11.1/2
a=5.55 ans.
_______________________________________________
I'll leave the rest for you to practice with.
4.9x^2+48x+64
5.45-80y^2
6. 15a^2-31ab+10b^2

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Factor:
1.28b+2ay+bx+xy
I think you have posted this incorrectly.
--------------------------------------------
2. y^2-7y-18
(y-9)(y+2)
-------------------------------------------
3. Y^2-13y+40
(y-5)(y-8)
--------------------------------------------
4.x^2-x-30
(x-6)(x+5)
------------------------------------------
5. 8x^2-14x-15
8x^2 -20x + 6x - 15
4x(2x -5) + 3(2x-5)
(2x-5)(4x+3)
-----------------------------------
6. 12a^2+81-21
12a^2+60
12(a^2+5)
-----------------------------------
7.3a(a-4)-(4-a)
= 3a(a-4)+(a-4)
= (a-4)(3a+1)
=======================
Cheers,
Stan H.

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the basic formula for solving an inequality is:
if |x| = d, then x = d, and x = -d
if |x| < d, then x < d, and x > -d
if |x| > d, then x > d, and x < -d
likewise,
if |x| <= d, then x <= d, and x >= -d
if |x| >= d, then x >= d, and x >= -d
-----
if |2x+1| <= 11, then
2x + 1 <= 11, and 2x + 1 >= -11
-----
working with 2x + 1 <= 11:
2x + 1 <= 11
2x <= 10
x <= 5
-----
working with 2x + 1 >= -11:
2x + 1 >= -11
2x >= -12
x >= -6
-----
answer is x <= 5 and x >= -6
this translates to:
-6 <= x <= 5
-----
to test:
let x = -7
|2x+1| = |-14+1| = |-13| = 13 NOT <= 11 = ok since -7 is not >= -6 which violates one of the conditions.
let x = 6
|2x+1| = |12+1| = |13| = 13 NOT <= 11 = ok since 6 is not <= 5 which violates one of the conditions.
let x = -6
|2x+1| = |-12+1| = |-11| = 11 <= 11 = ok since -6 = -6 which is one of the conditions.
let x = 5
|2x+1| = |10+1| = |11| = 11 <= 11 = ok since 5 = 5 which is one of the conditions.
let x = 0
|2x+1| = |-+1| = |1| = 1 <= 11 = ok since 1 > -6 and < 5 which is one of the conditions.
-----
answer proves to be good.
|2x-1| <= 11 is a true equation when:
-6 <= x <= 5
-----

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First question: 1/x^2-36}-{x+3/x^2-7x+6
-
Note that the 1st denominator is the "difference of squares"
2nd denominator can be factored also;
-
The common denominator is(x-6)(x+6)(x-1)
we have:
= = ; removing brackets changes signs
Combining like terms;
= =

Second question: 16 minus 1 over y^2 over 4 plus 1 over y

----------

:
We want each written as a single fraction:

----------

:
Invert the dividing fraction and multiply
*
:
16y^2 - 1 is the difference of squares, factor that to
*
;
Some simple canceling and you have:

:
:
"1/x^2-36 is a fraction as is x+3/x^2-7x+6 I am to subtract and then simplify if possible.
-
;
This is exactly the same as the 1st problem

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Ans: 1/3 and -1/3
=======================
Cheers,
Stan H.

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classify the polynomial as constant linear quadratic cubic or quartic and determine the leading term the leading coefficient and the degree of the polynomial f(x)=9x^2-12+0.17x-8x^3
----------
Type: cubic
Leading term: -8x^3
Leading coefficient: -8
Degree: three
====================
Cheers,
Stan H.

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Factor completely:

Swap the last two terms:



Factor y out of the first two terms:



Factor 1 out of the last two terms:



Factor out 



Factor 



Edwin

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We need to simplify this equation
3a^2-9a-12/6a^2+30a+24
--------------
Divide by 3 and factor out 2 in the DEN. That makes it easier to factor further.

=
=
It is simpler.
BTW, it's not an equation, it's an expression. An equation will have an equal sign, something equals something else.

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Start with the given expression


Factor the numerator


Factor the denominator


Highlight the common terms.


Cancel out the common


Simplify


Reduce


Distribute


So simplifies to


In other words, where or

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It is very hard to follow this question and even if I took what you have inputed precisely as it is, I do not think it would help you. If you send me an email with the question, then I will answer it for you. I think the best way to do this might be in a MS Paint file. Just draw the equation (doesn't have to be too neat, just readable). send file to justin.sheppard.tech@hotmail.com
.
I hope this helps!
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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Factor each term and then determine what factors are in each term.
1*x*x*x*x*x*y*y*y*y*y
1*x*x*x*x *y*y*y
1*x*x*x*x*y*y*y*y
-1*x*y
.
Because the last term only has one x and one y. The GCF is
.
I hope this helps!
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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You can factor out 2z^2 from each term and you are left with
.
*If you have trouble finding out if anything else will factor you can write out a factor tree for each term.
.
Because there are no common terms to all of the terms then there is nothing left to factor.
.
I hope this helps!
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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The only thing you can do on this one is factor out a 2 from each term.
.

.
I hope this helps!
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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This will not factor any further.
.
Private tutoring is available. Click on my name to go to my website or email me at justin.sheppard.tech@hotmail.com for more information. If you have any other questions you can direct them to me personally and I will answer them for you.

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# 1



Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,3,6,7,9,14,18,21,42,63,126
-1,-2,-3,-6,-7,-9,-14,-18,-21,-42,-63,-126


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*(-126)
2*(-63)
3*(-42)
6*(-21)
7*(-18)
9*(-14)
(-1)*(126)
(-2)*(63)
(-3)*(42)
(-6)*(21)
(-7)*(18)
(-9)*(14)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	-126	1+(-126)=-125
2	-63	2+(-63)=-61
3	-42	3+(-42)=-39
6	-21	6+(-21)=-15
7	-18	7+(-18)=-11
9	-14	9+(-14)=-5
-1	126	-1+126=125
-2	63	-2+63=61
-3	42	-3+42=39
-6	21	-6+21=15
-7	18	-7+18=11
-9	14	-9+14=5

From the table, we can see that there are NO pairs of numbers which add to .


So CANNOT be factored.

---

Start with the given expression


Factor out the GCF


Since the inner term contains no other common terms or exponents, this is as far as we go.


So factors to

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since P=2L+2W
we know value of P and W ....so plug them in
6x+6=2L+2(x)
4x+6=2L
this is the length of the given problem

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(x+1/7)(x-4/5)=0
X^2+X/7-4X/5-4/35=0
X^2+(5X-7*4X)/35-4/35=0
X^2+(5X-28X)/35-4/35=0
X^2-23X/35-4/35=0 ans.

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it can be written various ways but if is the amount of salt and
is the speed of the stream
then where k is a constant or it can be written as

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{1/x^2-36}-{x+3/x^2-7x+6}
= [1/(x-6)(x+6)] - [(x+3)/(x-3)(x-2)]
lcd = (x-6)(x+6)(x-3)(x-2)
Rewrite each fraction with lcd as its denominator:
= [(x-3)(x-2]/lcd - [(x+3)(x+6)(x-6)]/lcd
Combine the numerators over the lcd:
= [x^2-5x+6-(x+3)(x^2-36)]/lcd
= [x^2-5x+6-x^3-36x+3x^2-108]/lcd
= [-x^3+4x^2-41x-102]/lcd
================================
Cheers,
Stan H.

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xy-3x+4y-12/ xy+5x+4y+20
[x(y-3) + 4(y-3)] / [x(y+5) + 4(y+5)]
[(x+4)(y-3)] / [(x+4)(y+5)]
Cancel the common x+4 factor to get:
= (y-3)/(y+5)
=======================
xy+6x-3y-18/ xy+5x-3y-15
[x(y+6) -3(y+6)] / [x(y+5) -3(y+5)]
[(x-3)(y+6)] / [(x-3)(y+5)]
= (y+6)/(y+5)
=====================
Cheers,
Stan H.

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+ =
:
Factor where we can:
+ =
:
Multiply equation by (a+3)(x+1)
(a+3)(a+1)* + (a+3)(a+1)* = (a+3)(a+1)*
:
Cancel out the denominators and you have:
(3a-5) + 2(a+1)(a+1) = (a+3)(a-3)
:
FOIL
(3a - 5) + 2(a^2 + 2a + 1) = a^2 - 9
:
3a - 5 + 2a^2 + 4a + 2 = a^2 - 9
:
Combine like terms on the left:
2a^2 - a^2 + 3a + 4a - 5 + 2 + 9 = 0
:
a^2 + 7a + 6 = 0
:
Factor this to:
(a + 6)(a + 1) = 0
:
a = -6
and
a = -1
:
Check solution of x=-6 in original equation
+ =
Do the math, find the common denominator, and you have;
+ =
:
x=-1 can not be a solution, note that in the last denominator we have division by 0

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Given eqn,

Remember quadratic function --->
To have a clearer picture of the graph,
We first find the x-coordinate of the vertex with the formula:
---->
, x-coordinate of vertex
.
Then for the y-coordinate:

, y-coordinate of vertex
VERTEX -----> (-2,-49)
.
Now for the y-intercept: let x=0 --->
Solving or the x-intercept, via our eqn:
, factor out, perfect square

x-intercepts----
See graph below,
----> ALL INTERCEPTS are marked: GREEN circle=y-intercept and BLUE circle=x-intercepts with BLACK circle as VERTEX (-2,-49)
.
Thank you,
Jojo

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Factorise each term
Cancel out common terms

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Since varies inversely to , it follows:

where----
Continuing,


---> 6 hrs & 52.5 mins
*It make sense for to go up, since goes down,Varies Inversely.
Thank you,
Jojo

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x^2 + 3xy - 10y^2
(x+5y)(x-2y)

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Rachel allows herself 1 hour to reach a sales appointment 50 miles away. After she has driven 30 miles, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30 miles?
.
Let x = speed driven for first 30 miles
x+15 = speed driven for last 20 miles
.
And,
Let y = time driving first 30 miles
1-y = time driving for last 20 miles
.
First 30 miles (equation 1):
xy = 30
Last 20 miles (equation 2):
(x+15)(1-y) = 20
.
Solve equation 1 for y:
xy = 30
y = 30/x
.
Substitute the above into equation 2 and solve for x:
(x+15)(1-y) = 20
(x+15)(1-(30/x)) = 20
Multiplying both sides by x:
(x+15)(x-30) = 20x
FOIL the left side:
x^2-30x+15x-450 = 20x
x^2-15x-450 = 20x
x^2-35x-450 = 0
(x+10)(x-45) = 0
x = {-10, 45}
.
We can throw out the negative solution leaving us with:
x = 45 mph

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Please solve the problem. The product of two consecutive integers is 5 more than their sum. Find the integers.
.
Let x = first consecutive integer
then
x+1 = second consecutive integer
.
x(x+1) = 5 + x + x+1
x^2+x = 6 + 2x
x^2-x-6 = 0
(x-3)(x+2) = 0
x = {-2, 3}
.
Two possible solutions:
-2 and -1
3 and 4

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The length of a rectangular frame is 6 cm more than the width. The area inside the frame is 55 square cm. Find the width of the frame.
.
Let w = width of rectangular frame
then
w+6 = length of rectangular frame
.
Area = width * length
55 = w(w+6)
55 = w^2+6w
0 = w^2+6w-55
0 = (w+11)(w-5)
w = {-11, 5}
.
We can toss out the negative solution leaving us with:
w = 5 cm (width)

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The x-intercepts are found by setting y equal to zero and solving for 'x'.
y = x^2 + 4x - 45
0 = x^2 + 4x - 45
0 = (x+9)(x-5)
x = {-9, 5}
.
So, the x-intercepts are at (-9,0) and (5,0)

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By the moment that the object hits the ground, h=0.
Thus, substituting,
0=80t-16t²
factor out t
t(80-16t)=0
t=0 or 80-16t=0
16t=80
t=5
SO, it's 0 or 5 seconds.
HOwever, at t=0, the object is not falling but was about to go up with the initial velocity of 80 ft/sec
So, the answer is that t=5.

Power up,
HyperBrain!

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Find the polynomial for the perimeter of
6, 6x, 3, 9x+1, 3, 4x
:
Perimeter is the sum of all the sides:
P = 6 + 6x + 3 + (9x+1) + 3 + 4x
Add like terms:
P = 6x + 9x + 4x + 6 + 3 + 1 + 3
:
P = 19x + 13

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The polynomial is prime since there is no middle term and it is not a difference of squares.

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Start with the given expression


Group "a+b" in a parenthesis


Let


Replace each "a+b" with "z"


Factor out the GCF "z"


Plug in



So factors to

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Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,3,4,6,7,12,14,21,28,42,84
-1,-2,-3,-4,-6,-7,-12,-14,-21,-28,-42,-84


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*(-84)
2*(-42)
3*(-28)
4*(-21)
6*(-14)
7*(-12)
(-1)*(84)
(-2)*(42)
(-3)*(28)
(-4)*(21)
(-6)*(14)
(-7)*(12)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	-84	1+(-84)=-83
2	-42	2+(-42)=-40
3	-28	3+(-28)=-25
4	-21	4+(-21)=-17
6	-14	6+(-14)=-8
7	-12	7+(-12)=-5
-1	84	-1+84=83
-2	42	-2+42=40
-3	28	-3+28=25
-4	21	-4+21=17
-6	14	-6+14=8
-7	12	-7+12=5

From the table, we can see that the two numbers and add to (the middle coefficient).


So the two numbers and both multiply to and add to


Now replace the middle term with . Remember, and add to . So this shows us that .


Replace the second term with .


Group the terms into two pairs.


Factor out the GCF from the first group.


Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


Combine like terms. Or factor out the common term

---------------------------------------------


Answer:


So factors to .


Note: you can check the answer by FOILing to get or by graphing the original expression and the answer (the two graphs should be identical).

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Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,4
-1,-2,-4


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*4
2*2
(-1)*(-4)
(-2)*(-2)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	4	1+4=5
2	2	2+2=4
-1	-4	-1+(-4)=-5
-2	-2	-2+(-2)=-4

From the table, we can see that the two numbers and add to (the middle coefficient).


So the two numbers and both multiply to and add to


Now replace the middle term with . Remember, and add to . So this shows us that .


Replace the second term with .


Group the terms into two pairs.


Factor out the GCF from the first group.


Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


Combine like terms. Or factor out the common term

---------------------------------------------


Answer:


So factors to .


Note: you can check the answer by FOILing to get or by graphing the original expression and the answer (the two graphs should be identical).

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Start with the given expression


Factor out the GCF

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Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,3,5,6,9,10,15,18,30,45,90
-1,-2,-3,-5,-6,-9,-10,-15,-18,-30,-45,-90


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*(-90)
2*(-45)
3*(-30)
5*(-18)
6*(-15)
9*(-10)
(-1)*(90)
(-2)*(45)
(-3)*(30)
(-5)*(18)
(-6)*(15)
(-9)*(10)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number	Second Number	Sum
1	-90	1+(-90)=-89
2	-45	2+(-45)=-43
3	-30	3+(-30)=-27
5	-18	5+(-18)=-13
6	-15	6+(-15)=-9
9	-10	9+(-10)=-1
-1	90	-1+90=89
-2	45	-2+45=43
-3	30	-3+30=27
-5	18	-5+18=13
-6	15	-6+15=9
-9	10	-9+10=1

From the table, we can see that the two numbers and add to (the middle coefficient).


So the two numbers and both multiply to and add to


Now replace the middle term with . Remember, and add to . So this shows us that .


Replace the second term with .


Group the terms into two pairs.


Factor out the GCF from the first group.


Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


Combine like terms. Or factor out the common term

---------------------------------------------


Answer:


So factors to .


Note: you can check the answer by FOILing to get or by graphing the original expression and the answer (the two graphs should be identical).