Trinomials and binomials?

1) You can factor out an "a" from each term, and nother else.

a(8p^2 + 34p + 35)

Then, factor the inside trinomial.

a(4p + 7)(2p + 5)

2) Correct, it does not factor.

3) Factor out a 6r^2 first. (you can't factor 9 out of 480)

6r^2(9r^2 - 54r + 80)

6r^2(3r - 10)(3r - 8)

4) 2a^3 +16. Yes, you can only factor out a 2. 2(a^3 + 8)

5) 2x^2+4= -36

Subtract (don't add) 4 from both sides to get:

2x^2 = -40

Now, divide by 2 to get: x^2 = -20

Take the square root of both sides to get:

√x^2 = √-20

x = 2√-5

When you have a negative under a square root, you need the complex number i.

x = 2i√5

8p^2a + 34pa+ 35a

= a(8p^2 + 34p+ 35)

= a(8p^2 + 20p + 14p + 35)

= a[4p(2p + 5) + 7(2p + 5)]

= a[(4p + 7)(2p + 5)]

54r^4 - 324r^3 + 480r^2

= 6a²(9r^2 - 54r + 80)

= 6a²(9r^2 - 24r – 30r + 80)

= 6a²[3r(3r - 8) – 10(3r – 8)]

= 6a²[(3r – 10)(3r – 8)]

2a³ + 16

= 2(a³ + 8)

= 2(a + 2)(a² – 2a + 4)

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