SOLUTION: Factor completely: x^2 - 5x - 7x + 35 = x^2 - 12x + 35 2x^2 + 11x + 5 x^4y - 16y = x^4 - 15y Thank you so much!

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Factor completely: x^2 - 5x - 7x + 35 = x^2 - 12x + 35 2x^2 + 11x + 5 x^4y - 16y = x^4 - 15y Thank you so much!      Log On


   



Question 196046: Factor completely:
x^2 - 5x - 7x + 35 = x^2 - 12x + 35
2x^2 + 11x + 5
x^4y - 16y = x^4 - 15y
Thank you so much!

Found 2 solutions by jim_thompson5910, RAY100:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that you want to factor x%5E2+-+12x+%2B+35




Looking at the expression x%5E2-12x%2B35, we can see that the first coefficient is 1, the second coefficient is -12, and the last term is 35.


Now multiply the first coefficient 1 by the last term 35 to get %281%29%2835%29=35.


Now the question is: what two whole numbers multiply to 35 (the previous product) and add to the second coefficient -12?


To find these two numbers, we need to list all of the factors of 35 (the previous product).


Factors of 35:
1,5,7,35
-1,-5,-7,-35


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


These factors pair up and multiply to 35.
1*35
5*7
(-1)*(-35)
(-5)*(-7)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -12:


First NumberSecond NumberSum
1351+35=36
575+7=12
-1-35-1+(-35)=-36
-5-7-5+(-7)=-12



From the table, we can see that the two numbers -5 and -7 add to -12 (the middle coefficient).


So the two numbers -5 and -7 both multiply to 35 and add to -12


Now replace the middle term -12x with -5x-7x. Remember, -5 and -7 add to -12. So this shows us that -5x-7x=-12x.


x%5E2%2Bhighlight%28-5x-7x%29%2B35 Replace the second term -12x with -5x-7x.


%28x%5E2-5x%29%2B%28-7x%2B35%29 Group the terms into two pairs.


x%28x-5%29%2B%28-7x%2B35%29 Factor out the GCF x from the first group.


x%28x-5%29-7%28x-5%29 Factor out 7 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.


%28x-7%29%28x-5%29 Combine like terms. Or factor out the common term x-5

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


Answer:


So x%5E2-12x%2B35 factors to %28x-7%29%28x-5%29.


Note: you can check the answer by FOILing %28x-7%29%28x-5%29 to get x%5E2-12x%2B35 or by graphing the original expression and the answer (the two graphs should be identical).








# 2




Looking at the expression 2x%5E2%2B11x%2B5, we can see that the first coefficient is 2, the second coefficient is 11, and the last term is 5.


Now multiply the first coefficient 2 by the last term 5 to get %282%29%285%29=10.


Now the question is: what two whole numbers multiply to 10 (the previous product) and add to the second coefficient 11?


To find these two numbers, we need to list all of the factors of 10 (the previous product).


Factors of 10:
1,2,5,10
-1,-2,-5,-10


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


These factors pair up and multiply to 10.
1*10
2*5
(-1)*(-10)
(-2)*(-5)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 11:


First NumberSecond NumberSum
1101+10=11
252+5=7
-1-10-1+(-10)=-11
-2-5-2+(-5)=-7



From the table, we can see that the two numbers 1 and 10 add to 11 (the middle coefficient).


So the two numbers 1 and 10 both multiply to 10 and add to 11


Now replace the middle term 11x with x%2B10x. Remember, 1 and 10 add to 11. So this shows us that x%2B10x=11x.


2x%5E2%2Bhighlight%28x%2B10x%29%2B5 Replace the second term 11x with x%2B10x.


%282x%5E2%2Bx%29%2B%2810x%2B5%29 Group the terms into two pairs.


x%282x%2B1%29%2B%2810x%2B5%29 Factor out the GCF x from the first group.


x%282x%2B1%29%2B5%282x%2B1%29 Factor out 5 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.


%28x%2B5%29%282x%2B1%29 Combine like terms. Or factor out the common term 2x%2B1

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


Answer:


So 2x%5E2%2B11x%2B5 factors to %28x%2B5%29%282x%2B1%29.


Note: you can check the answer by FOILing %28x%2B5%29%282x%2B1%29 to get 2x%5E2%2B11x%2B5 or by graphing the original expression and the answer (the two graphs should be identical).








# 3

I'm assuming you want to factor x%5E4y+-+16y


x%5E4y+-+16y Start with the given expression


y%28x%5E4+-+16%29 Factor out the GCF


y%28x%5E2+-+4%29%28x%5E2+%2B4%29 Factor x%5E4+-+16 using the difference of squares formula


y%28x-2%29%28x%2B2%29%28x%5E2+%2B4%29 Factor x%5E2+-+4 using the difference of squares formula


So x%5E4y+-+16y completely factors to y%28x-2%29%28x%2B2%29%28x%5E2+%2B4%29

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
(1) x^2 -12x +35
(x-5)(x-7)
.
(2) 2X^2 +11x +5
(2x+1)(x+5)
.
(3) x^4y -16y = x^4 -15y
x^4y -16y +15y -x^4 =0
x^4y -1y -x^4 =0
x^4 (y-1) -y =0, easy answer
.
or
x^4y -16y+15y -x^4 =0
x^4y -1y -x^4 =0
y(x^4 -1) -x^4 =0
y(x^2 +1)(x^2-1) -x^4 =0
y(x^2 +1)(x+1)(x-1) -x^4 =0,,,,better answer
.
checking
Let x=1, y=2,
original is (-1), answer is (-1) ,,,,ok