Question

Find factors if y2+10y+4

Answer 1

Y2+10y+4
D=b2-4ac
D=100-16
D=84
Root exist D>0
Now
Y= -b+rootD/2a
Y= -10+root84/2
And
Y=. -b-rootD/2a
Y= -10-root84/2

Answer 2

Solution:

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Given:


To Factorize:

y² + 10y + 4

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As we know,.

The given polynomial is of the form,

ax² + bx + c = 0,

So,



We know that,

(a + b)² = a²+ 2ab + b²,..

so,

attempting to bring this equation to (a+b)² form,.

we get,

=> y² + 10y + 4 = 0,

=> $y^2 + 2( \frac{10}{2})y + 4 = 0$  (Multiplying and diving 10y by 2)

=> $y^2 + 2( \frac{10}{2})y + ( \frac{10}{2})^2 + 4 = 0 + (\frac{10}{2})^2$

(adding $( \frac{10}{2})^2$ on both the sides )

=> $(y+ \frac{10}{2})^2 + 4 = (5)^2$

=> $(y + 5)^2 + 4 = 25$

=> $(y+5)^2 = 25 - 4$

=> $(y+5)^2 = 21$

=> $y + 5 = \sqrt{21}$

=> $y = \sqrt{21} - 5$  (or)

     $y =- 5 - \sqrt{21}$

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We know that,

 subtracting the number by itself = 0

y = -5 + √21 or y = -5 - √21

so,

=> (y - y) (y - y) = 0,

=> (y -(- 5 + √21))  (y - (-5 -√21 ) ) = 0,

=> (y + 5 - √21) (y +5 +√21) = 0,.

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                             Hope it Helps!!