Could you please explain this to me? Factor each polynomial completely...2x^2 + 9x +10?

Question

OK, I know that 2x^2 + 9x +10 = (2x^2 + 5)(x + 2), but I do not understand how to do the following...

a) x^4 - 16

b) x^2 - 28x + 196

AND the one that just has me COMPLETELY stumped...

c) 4x^3 - 44x^2 + 112x

I just do not know HOW to factor them! HELP PLEASE!

Answer 1

a) a difference of squares so it factors into a sum and difference
(x^2-4)(x^2+4)
x^2 + 4 does not factor over the real numbers and is done, but (x^2-4) is another difference of squares and factors into (x-2)(x+2)
Final answer (x-2)(x+2)(x^2+4)

b. this is called a perfect square trinomial and it factors into 2 binomials which are identical. Since the x term is negative both binomials will be differences. The square root of 196 is 14. The answer is (x-14)(x-14) or (x-14)^2

c. First factor out the GCF which is 4x
4x(x^2 -11x + 28)
Factor the quadratic
4x(x-7)(x-4)

Answer 2

put y = x^2
then y^2 - 16 = 0
=> (y+2)(y-2) = 0
=> y = +- 2
=> x = +-2
=> x = +-sqrt(2) and the other two roots are imaginary - +-sqrt(2) * i

b) (x - 14)^2 = 0
=> x = 14

c) 4x * ( x^2 - 11x + 28 ) = 0
=> 4x * ( x - 7 ) ( x - 4 ) = 0
=> x =0, 4, 7

Answer 3

2x² + 9x + 10 = (2x + 5).(x + 2)

a) (x² - 4).(x²+ 4) = (x - 2).(x + 2).(x² + 4)

b) (x - 14).(x - 14) = (x - 14)²

c) 4x.(x² - 11x + 28) = 4x.(x - 7).(x - 4)

Answer 4

a) x^4-16
=(x^2)^2-(4)^2
=(x^2+4)(x^2-4) [a^2-b^2=(a+b)(a-b) formula applied]
=(X^2+4){(x)^2-(2)^2}
=(x^2+4)(x+2)(x-2)
b) x^2-28x+196
=(x)^2-2*x*14+(14)^2
=(x-14)^2 [a^2-2ab+b^2=(a-b)^2 formula applied]
c)4x^3-44x^2+112x
=4x(x^2-11x+28x)
=4x(x^2-7x-4x+28)
=4x{x(x-7)-4(x-7)}
=4x(x-7)(x-4)