Question

factorise : x2-10x+21

Answer 1

Answer:

(x-7) (x-3)

Step-by-step explanation:

We have to factorise the quadratic equation x²-10x+21 .

As the Highest power of this equation is 2, then its highest number of factors will be 2.

We can compare this equation with the quadratic equation

$a {x}^{2} + bx + c$

Here, a=1, b =-10 , c=21

We have to find out two numbers such that if we multiply them we should get 21 and if we add them we should get -10 .

Such two numbers are -7 and -3.

So we can write the given quadratic equation as

${x}^{2} + ( - 7 - 3)x + 21$

If we open the bracket, we get,

${x}^{2} - 7x - 3x + 21$

we can see that, in the first two terms i.e. x² and -7x, x is common and -3 is common in next two terms i.e. -3x and 21.

So by taking x common from first two terms and -3 common from next two terms, we get,

$x(x - 7) - 3(x - 7)$

Now we have two terms and (x-7) is common in both terms. So if we take (x-7) common from this two terms, we get,

$(x - 7)(x - 3)$

Conclusion:

If we factorise x²-10x+21 , we get (x-7)(x-3).