Question

factorise 9x^2+6x+1​

Answer 1

$\huge \mathfrak \red{answer}$

$\bf\pink{ \implies \: = (3x - 1)(3x - 1)}$

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$\sf \underline \pink{Question}$

$\rm{9 {x}^{2} - 6x + 1}$

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$\sf \underline \orange{step \: by \: step \: explanation}$

$\tt \red{ \implies \: 9 {x}^{2} - 6x + 1}$

$\tt \green{ \implies \: = 9 {x}^{2} - 3x - 3x + 1}$

$\tt \blue{ \implies \: = 3x(3x - 1) - 1(3x - 1)}$

$\tt \purple{ \implies \: = (3x - 1)(3x - 1)}$

Answer 2

Answer:

(3x+1)(3x+1) or (3x+1)^2

Step-by-step explanation:

We know that we have to split the middle term that is the term [constant×X] in such a way that the new constant have product equal to product of coefficient of a and the last term that is constant.

=>if question is of form ax^2+bx+c

then split bx into-[dx + ex] such that d×e=a×c

So let's apply it on 9x^2+6x+1

here we have to break 6 so that the product comes 9×1 that is 9.

Let's find factors of 9 which add upto 6.

We find 9=3×3

and,3+3=6

=> 9x^2+6x+1 = 9x^2+3x+3x+1

=>take two terms and remove some common part from them

=>9x^2 is 3x × 3x

so take 3x common from 9x^2+3x

=>3x(3x+1)+3x+1

now take 1 common from 3x+1 which was left.

=>3x(3x+1)+1(3x+1)

now two terms have 3x+1 take this common

=>(3x+1)(3x+1) or (3x+1)^2

Hence factorised.

If you need the values of x that is the x where 9x^2+6x+1 is equal to zero.

Then (3x+1)(3x+1)=0

=>(3x+1)^2=0

=>3x+1=0

=>3x=(-1)

=>x=(-1)/3

Hope it helps...

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