Olympiad Test: Factorisation

ax+by+bx+az+ay+bz, rearranging we get ax+ay+az+bx+by+bz
a(x+y+z)+b(x+y+z)=(a+b)(x+y+z). Hence the factors are (a+b), (x+y+z)

x⁴ + 4

= (x² + 2)² - 4x²

= (x² + 2)² - (2x)²

a² - b² = (a + b)(a - b)

= (x² + 2 + 2x)(x² + 2 - 2x)

Given equation is x⁴ + 2x² + 9
We can rewrite this as,
(x²)² + 6x² + 9 − 4x²
⇒ (x² + 3)² − (2x)²
....Since a² + 2ab + b² = (a+b)²
⇒ x⁴ + 2x² + 9 = (x² − 2x + 3)(x² + 2x + 3)     
....Since a² − b² = (a+b)(a−b)

Let the other factor be x²+ax+b. We have (x²+2x+5)(x²+ax+b)
= x⁴+px²+q                
x⁴(2+a)x³(2a+b+5)x²(5a+2b)x+5b = x4+px²+q
Comparing the coefficients of corresponding terms, we get 2a+b+5 = p  ......(1)
5b = q   ......(2)                
2+a = 0 ⇒ a =−2 
5a+2b = 0 ⇒ b = 5
∴ p = 2a+b+5 = 2(−2)+5+5 = 6
q = 5b = 5(5) = 25

x² + xy - 2xz - 2yz

x(x + y).-2z(x + y)

(x - 2z)(x + y)

(x−5)² = x²−10x+25 using 

(a−b)² = a²−2ab+b². So, Pankaj's answer is correct.

(a−4)(a−2) = a²−4a−2a+8 = a²−6a+8 So, the statement in option (A) is correct.

x⁴+y⁴+x²y² =(x⁴+y⁴+2x²y²)−x²y² =(x²+y²)²−(xy)²
=(x²+y²+xy)(x²+y²−xy)

15x²−26x+8 =15x²−20x−6x+8 =5x(3x−4)−2(3x−4) =(3x−4)(5x−2)

f(x)=x9−x
=x(x8−1)
=x[(x4)2−(1)2]
=x[(x4−1)(x4+1)]
=x(x4+1)[(x2)2−12]
=x(x4+1)(x2+1)(x2−1)
=x(x4+1)(x2+1)(x−1)(x+1)
∴ There are 5 factors of f(x)⇒(x),(x4+1),(x2+1),(x−1) and (x+1)

a² - b + ab - a = a² + ab - b - a

= ( a² + ab ) - ( b + a )

= a ( a + b) - (a + b )

= ( a + b) ( a - 1 )

p^2 + 7p + 10 / p + 5

p^2 + 5p + 2p + 10 / p+5

p ( p + 5 ) + 2(p + 5 ) / p+5

( p+5) ( p+ 2 ) / p + 5

( p+2) 

n(3n+2) = 3n²+2n

(x²+3x+5)(x²−3x+5) = (x²+5+3x)(x²+5−3x) = (x²+5)²−(3x)² = m²−n²
∴ m = x²+5

b² - 7b+12=b²-3b-4b+12=b(b-3)-4(b-3)=(b-3)(b-4)

6mn-9m-4n+6  

take 3m common  

3m(2n-3)-2(2n-3)  

(3m-2)(2n-3)