Math, asked by kolisulbha473, 7 months ago

factorise the following x²+6x+9 and x²+2x-15 and 49a²+70ab+25b² and 25x²-49y² and x²+6x+9​

Answers

Answered by pawansweta5
0

Answer:

If y = x^2 - 6x + 9, what is the value of x?

(1) y = 0

(2) x + y = 3

Answered by veer0407
0

Step-by-step explanation:

1

We can Split the Middle Term of this expression to factorise it

In this technique, if we have to factorise an expression like ax2+b+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅9=9

AND

N1+N2=b=6

After trying out a few numbers we get N1=3 and N2=3

3⋅3=9, and 3+3=6

x2+6x+9

=x2+3x+3x+9

=x(x+3)+3(x+3)

=(x+3)(x+3)

2

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like ax2+bx+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅(−15)=−15

AND

N1+N2=b=2

After trying out a few numbers we get N1=5 and N2=−3

5⋅(−3)=−15, and 5+(−3)=2

x2+2x−15=x2+5x−3x−15

=x(x+5)−3(x+5)

=(x−3)(x+5) .

(x−3)(x+5) is the factorised form of the expression.

3

The factors of 49a^{2}+70ab+25b^{2}49a2+70ab+25b2 is (7a+5b)and (7a+5b)

Solution:

Given  49a^{2}+70ab+25b^{2}49a2+70ab+25b2 ,

49=7^{2}49=72 and 25=5^{2}25=52

On expanding the equation we get,

\Rightarrow 7\times7\times a^{2}+70ab+ 5\times5\times b^{2}⇒7×7×a2+70ab+5×5×b2

\Rightarrow (7a)^{2}+70ab+(5b)^{2}⇒(7a)2+70ab+(5b)2

70ab can be written as 2\times5\times7\times ab2×5×7×ab

Therefore, \Rightarrow (7a)^{2}+2\times5\times7\times ab+(5b)^{2}⇒(7a)2+2×5×7×ab+(5b)2

It is in the form of a^{2}+2ab+b^{2} = (a+b)^{2}a2+2ab+b2=(a+b)2

Here, a is 7a and b is 5b

On substituting the values we get,

(7a\times5b)^{2}(7a×5b)2 or (7a\times5b)(7a\times5b)(7a×5b)(7a×5b)

Hence, the factors are (7a+5b)and (7a+5b).

4

25x^2-49y^2

5x^2-7y^2(a^2-b^2=(a-b)(a+b))

(5x-7y)(5x+7y)

5

We can Split the Middle Term of this expression to factorise it

In this technique, if we have to factorise an expression like ax2+b+c, we need to think of 2 numbers such that:

N1⋅N2=a⋅c=1⋅9=9

AND

N1+N2=b=6

After trying out a few numbers we get N1=3 and N2=3

3⋅3=9, and 3+3=6

x2+6x+9

=x2+3x+3x+9

=x(x+3)+3(x+3)

=(x+3)(x+3)

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