Practice questions for Factorization

(i) p² + 6p + 9= (p+3)²
(ii) 4z² – 4z + 1=(2z+1)²
(iii) 4p⁴ + 9p⁴ + 12p²q²=(2p²+ 3q²)²
(iv) (a⁴ - 8a²b² + 16b⁴) - 81=(a²-4b²)²
(v) 625 – x² – 2xy – y²= 625 -(x²+ 2xy + y²) =(25-x-y)(25+x+y)

(i) 5(x+1)
(ii) 3(x+5)
(iii) 2(p+q)
(iv) 5(y² + 5y + 3)
(v) (x+1)(x+3)

(i) x(px+q)
(ii) 7(x²+3y²)
(iii) 2x(x²+y²+z²)
(iv) (a+b)(p²+q²)
(v) (n+1)(m+n)
(vi) (y-z)(y+16)
(vii) 5y² – 20y – 8z + 2yz=5y(y-4) +2z(y-4)=(y-4)(5y+2z)
(viii)10xy + 4x+ 5y + 2=2x(5y+2) + 1(5y+2)=(2x+1)(5y+2)
(ix) 6xy - 4y + 6 - 9x= 6xy-9x- 4y+6= 3x(2y-3)-2(2y-3)= (3x-2)(2y-3)

(i) 576 - p² + 8pq - 16q²= 24² -(p² - 8pq + 16q²)=24² -(p-4q)²=(24-p+4q)(24+p-4q)
(ii) 25x² - (y + z)² =(5x)²-(y + z)²=(5x-y-z)(5x+y+z)
(iii) 4p² - 4p - 3=4p² - 6p+2p - 3=2p(2p-3) +1(2p-3)=(2p+1)(2p-3)
(iv) 81 - x²- y²- 2xy= 9² -(x²+ y²+ 2xy)=9²-(x+y)²=(9-x-y)(9+x+y)
(v) 121 - x²- y² - 2xy= 11² -(x²+ y²+ 2xy)=11²-(x+y)²=(11-x-y)(11+x+y)

(i) 4a²- 12ab + 9b² =(2a-3b)²
(ii) 36p²- 84pq + 49q²=(6p-7q)²
(iii) 4a⁸- b⁸= (2a⁴)² - (b⁴)²=(a⁴ - b⁴)(a⁴+ b⁴) =(a²- b²)(a² + b²)(a⁴+ b⁴)
(iv) (z - 1)²+ 2 (z-1) (2z+3) + (2z +3)² =[(z-1)+(2z+3)]²=(3z+2)²
(v)p² -9q²=(p-3q)(p+3q)

Mean = Sum of Observation/Number of Obervations
Therefore
$M= \frac {x^4 -625}{x+5}= \frac { (x^2 +25)(x-5)(x+5)}{ x+5}=(x^2 +25)(x-5)$

Area of Triangle= (1/2) x Height X Base
Therefore
$A= \frac {1}{2} (x^4 + y^4) \times 14xy=7xy(x^4 + y^4)$

$p^4 -(p - q)^4 =[p^2 - (p-q)^2] [p^2 + (p-q)^2]=[p -(p-q)][p+(p-q)][p^2 + p^2 + q^2 -2pq]=q(2p-q)(2p^2 + q^2 -2pq)$

$A= \pi r^2$ or
$r^2 =\frac {A}{\pi} = \frac {\pi x^2 + 6\pi x + 9 \pi}{\pi}=x^2 + 6x + 9=(x+3)^2$
$r=(x+3)

$63p^2q^2 r^2 s – 9pq^2 r^2 s^2 + 15p^2qr^2 s^2 – 60p^2 q^2 rs^2= 3pqrs(21pqr - 3qrs + 5prs -20pqs)$