Question

the factories of 4y² +12y + 5​

Answer 1

Answer:

Given to factorise:-

4y² +12y + 5

Identity used:-

${x}^{2} +( a + b)x + ab = (x + a)( x+ b)$

Solution:-

4y² +12y + 5

Product=20

Sum= 12

Numbers= 10,2

Splitting the mid term:-

$4 {y}^{2} + 10y + 2y + 5$

$= 2y(2y + 5) + 1(2y + 5)$

$= (2y + 1)(2y + 5)$

Hence,The given solution:-

$(2y + 1)(2y + 5)$

Answer 2

$\huge\fbox\pink{Answer}$

$No.whose \:sum\: is \:12\: and\: product\: is\: 4\:\:\:×5=20 \:are\: 2\: and\: 10\: as\: 10+2=12 \:and\: 10×2=20$

$so 4y²+10y+2y+5$

$=2y(2y+5)+1(2y+5)$

$=(2y+5)(2y+1)$