How do you factor (16 y3 + 68 y2 + 42 y)?????

Answer 1

1) find a common factor in each term. In this case it would be 2y. that leaves you with 2y(8y2 + 34y + 21). 2y is one factor, now you just have to find out if 8y2 + 34y + 21 can be reduced further. in some cases you'll have to use the quadratic equation, but it's not needed for this one.

2) to reduce 8y2 + 34y + 21, start by finding multiplication factors for the first and last term. in this case there is only 1 possibility for each term. 8y2 can only be achieved by multiplying 4y and 2y. 21 can only be achieved by multiplying 7 and 3.

3) try out combinations of factors from the first and third terms until you get what you want. since there was only 1 possiblity for each term this makes it easy as there are only two possible combinations. (4y + 3)*(2y + 7) or (4y +7 )*(2y + 3).

4) for each combination multiply the outside terms and add them the product of the inside terms. for (4y + 3)*(2y + 7) the product of the outside terms is 28y (4y * 7), the product of the inside terms is 6y (2y *3). the sum of these is 34y. that's the combo were looking for. conversely, the other combo produces an outside product of 12y and an inside product of 14y. this gives 26y, but we know that's wrong.

5) write it all out and there's your answer: 2y(4y+3)(2y+7)

Answer 2

First you put 16 y in factor of the rest ... you get :
16 y ( y^2 + 68 y /16 + 42/16 ), which you can further factorize as a quadratic term.

You can factorize the quadratic term by remembering that y^2+2ay+a^2 = (y+a)^2
... let us say that y^2 + 68y/16 is the beginning of that term, 'a' is here equals to 34/16, and we are therefore missing (34/16)^2. Let us add and subtract that last term within the quadratic form ... it comes:
16y ( y^2 + 2(34/16)y +(34/16)^2 -(34/16)^2 +42/16 )
16y ( ( y + 34/16 )^2 -(34/16)^2 +42/16 )
Let us simplify the constants a little ...
-(34/16)^2 + 42/16= -34^2/16^2 + 42/16 = -17^2/(16*4) + 168/(16*4) = (-289+168)/(68) = -121/8^2 = -(11/8)^2

So we have 16y ( ( y + 34/16 )^2 - (11/8)^2 )
Now you have to remember that A^2-B^2 = (A-B)(A+B), which leads in our case to :
16y ( ( y + 34/16 ) - (11/8) )( ( y + 34/16 ) + (11/8) )
16y ( y + 34/16 - 11/8 )( y + 34/16 + 11/8 )
16y ( y + 17/8 - 11/8 )( y + 17/8 + 11/8 )
16y ( y + 6/8 )( y + 28/8 )
16y ( y + 3/4 )( y + 7/2 )

Answer 3

Factoring...

First we remove 2y from each term since it is the common term
2y(8y^2 + 34y + 21)
then we notice that 8 = 2 * 4, and 21 = 7 * 3
so we look for combinations of these factors that when multiplied across and added equal 34. We notice that 4*7 + 2*3 = 34.
2y(4y + 3)(2y + 7)

And this is your solution

Answer 4

16 y3 + 68 y2 + 42 y=y(16y^2+68y+42)=y(2y+7)(8y+6)
check
(2y+7)(8y+6)
16y^2+56y+12y+42
16y^2+68y+12

Answer 5

Ah ha! this might genuine be a chemistry question, so the respond ought to be Yttrium cubed enhanced by using a molecule of Yttrium cubed minus Yttrium equalling 27 enhanced by using 26 Yttria the situation cubed structures are authorized with valencies that don't interfere with 'conventional' section nor the interstitial fragility inherent in compounds whose commonality is very uncommon.

Answer 6

u take out y then see what the greatest common factor is which im not sure...maybe 4?

so you put in GCFy(#y2 + #y + #)

its very easy...once you practice a lot you will get a hang of it

Answer 7

16y^3 + 68y^2 + 42y
2y(8y^2 + 34y + 21)
2y(2y + 7)(4y + 3)

Answer 8

2y(8y2 + 34y + 21)
2y(4y + 2)(7y + 3)

Answer 9

y * (16y2 + 68y + 42)
y * (Ay + B) * (Cy + D)
Can you get A, B, C, and D?