Question
Factorise : $(a^2- 3a) (a^2 - 3a + 7) + 10$
✓
Answer
$(a^2- 3a) (a^2 - 3a + 7) + 10$
Let us assume $, a^2 - 3a = x$
Then, our polynomial becomes,
$( a^2 - 3a )( a^2 - 3a + 7 ) + 10$
$= x( x + 7 ) + 10$
$= x^2 + 7x + 10$
$= x^2 + 5x + 2x + 10$
$= x( x + 5 ) + 2 ( x + 5 )$
$= ( x + 5 )( x + 2 )$
By resubstituting the value of $x,$
$= (a^2 - 3a + 5)( a^2 - 3a + 2 )$
Now$, a^2 - 3a + 5$ will have no factor as discriminant is $-11$ that is less than $0$.
And,
$\therefore a^2 - 3a + 2 = a^2 - 2a - a + 2 = a( a - 2) - 1(a - 2) = (a - 1)(a - 2)$
So, factor of given polynomial are,
$a^2 - 3a + 2 = a^2 - 2a - a + 2$
$= (a^2 - 3a + 5)(a - 1)(a - 2)$
Let us assume $, a^2 - 3a = x$
Then, our polynomial becomes,
$( a^2 - 3a )( a^2 - 3a + 7 ) + 10$
$= x( x + 7 ) + 10$
$= x^2 + 7x + 10$
$= x^2 + 5x + 2x + 10$
$= x( x + 5 ) + 2 ( x + 5 )$
$= ( x + 5 )( x + 2 )$
By resubstituting the value of $x,$
$= (a^2 - 3a + 5)( a^2 - 3a + 2 )$
Now$, a^2 - 3a + 5$ will have no factor as discriminant is $-11$ that is less than $0$.
And,
$\therefore a^2 - 3a + 2 = a^2 - 2a - a + 2 = a( a - 2) - 1(a - 2) = (a - 1)(a - 2)$
So, factor of given polynomial are,
$a^2 - 3a + 2 = a^2 - 2a - a + 2$
$= (a^2 - 3a + 5)(a - 1)(a - 2)$
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