factorization (cross-method)

29.
-20a² - 95ab + 25b²
= -5(4a² + 19ab - 5b²)
= -5(4a - b)(a + 5b)


30.
3c²d² - 5cde - 2e²
= 3(cd)² - 5(cd)e - 2e²
= (3cd + e)(cd - 2e)


31.
-252 + 27x + 9x²
= 9(x² + 3x - 28)
= 9(x - 4)(x + 7)


32.
4p - 5q - 16p² + 40pq - 25q²
= (4p - 5q) - [16p² - 40pq + 25q²]
= (4p - 5q) - [(4p)² - 2(4p)(5q) + (5q)²]
= (4p - 5q) - (4p - 5q)²
= (4p - 5q)[1 - (4p - 5q)]
= (4p - 5q)(1 - 4p + 5q)


33.
2p⁴ - 5p²q² + 3q⁴
= 2(p²)² - 5(p²)(q²) + 3(q²)²
= (2p² - 3q²)(p² - q²)
= (2p² - 3q²)(p - q)(p + q)


34.
Put u = (2x + y)³
Then, x³ - (2x + y) ³
= x³ - u³
= [x - u] [x² + xu + u²]
= [x - (2x + y)] [x² + x(2x + y) + (2x + y)²]
= [x - 2x + y] [x² + 2x² + xy + 4x² + 4xy + y²]
= (y - x)(7x² + 5xy + y²)

29
-20a^2-95ab+25b^2
=5(a+5b)(4a-b)

30
3c^2d^2-5cde-2e^2
=(cd-2e)(3cd+1e)

31
-252+27x+9x^2
=(x-4)(x+7)

32
4p-5q-16p^2+40pq-25q^2
=-(16p^2-40pq+25q^2)+(4p-5q)
=-(4p-5q)^2+(4p-5q)
=(4p-5q)(1-4p+5q)

33
2p^4-5p^2q^2+3q^4
=(2p^2-3q^2)(p^2-q^2)
=(2p^2-3q^2)(p-q)(p+q)

34
x^3-(2x+y)^3
=[x-(2x+y)][x^2+x(2x+y)+(2x+y)^2]
=(3x+y)(x^2+2x^2+xy+4x^2+4xy+y^2)
=(3x+y)(7x^2+5xy+y^2)