Question
Factorise :$(a^2- 3a) (a^2 - 3a + 7) + 10$
✓
Answer
$\left(a^2-3 a\right)\left(a^2-3 a+7\right)+10$ Let us assume, $a^2-3 a=x$
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, $a^2-3 a+5$ 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,
$a^2 - 3a + 2 = a^2 - 2a - a + 2$
$= (a^2 - 3a + 5)(a - 1)(a - 2)$
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, $a^2-3 a+5$ 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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(a) 10 units (b) 8 units (c) 6 units (d) 12 units
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(a) square (b) rectangle (c) rhombus (d) kite
Q.5. PQRS is a :
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