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$${\mathtt{8}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{26}}{\mathtt{\,\times\,}}{\mathtt{y}}{\mathtt{\,\small\textbf+\,}}{\mathtt{15}}$$What is the best way to factor: 8y^2 +26y +15

Guest May 25, 2015

Best Answer 

 #1
avatar+981 
+10

The quadratic formula is ususally an option but here I would probably do this:

 

$$8y^2 + 26y + 15$$   

( Note that 8 does not go easily into the other terms, making factorising by straight up removing the 8 out of the question. Instead we can look at the product of 8 and 15, with the intention of splitting 25 up into 2 terms which can then be factorised. What 2 numbers add to make 26 and multiply to make (8*15) 120? 6 and 20)

 

$$8y^2 + 6y + 20y + 15$$

 

( Now we factorise in pairs)

 

$$2y(4y + 3) + 5(4y + 3)$$

 

$$(2y + 5)(4y + 3)$$

 

YAY :)

 

 

 

Just to check.....

 

$$(2y + 5)(4y + 3)$$

 

$$(2y\times{4y}) + (2y\times{3}) + (5\times{4y}) + (5\times{3})$$  (FOIL)

 

$$8y^2 + 6y + 20y + 15$$

 

$$8y^2 + 26y + 15$$

zacismyname  May 26, 2015
 #1
avatar+981 
+10
Best Answer

The quadratic formula is ususally an option but here I would probably do this:

 

$$8y^2 + 26y + 15$$   

( Note that 8 does not go easily into the other terms, making factorising by straight up removing the 8 out of the question. Instead we can look at the product of 8 and 15, with the intention of splitting 25 up into 2 terms which can then be factorised. What 2 numbers add to make 26 and multiply to make (8*15) 120? 6 and 20)

 

$$8y^2 + 6y + 20y + 15$$

 

( Now we factorise in pairs)

 

$$2y(4y + 3) + 5(4y + 3)$$

 

$$(2y + 5)(4y + 3)$$

 

YAY :)

 

 

 

Just to check.....

 

$$(2y + 5)(4y + 3)$$

 

$$(2y\times{4y}) + (2y\times{3}) + (5\times{4y}) + (5\times{3})$$  (FOIL)

 

$$8y^2 + 6y + 20y + 15$$

 

$$8y^2 + 26y + 15$$

zacismyname  May 26, 2015
 #2
avatar+92781 
+5

That is an excellent answer Zac  

 

Here is a video to reinforce what Zac has told you   

 

https://www.youtube.com/watch?v=ZQ-NRsWhOGI

Melody  May 26, 2015
 #3
avatar+981 
0

Thanks 

zacismyname  May 26, 2015

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