Please factor (3x^4)+(14x^3)+(23x^2)+(10)?

Question

(3x^4)+ (14x^3)+ (23x^2)+ (10)

Answer 1

3x^4 + 14x^3 + 23x^2 + 10

Let p(x) = 3x^4 + 14x^3 + 23x^2 + 10

By Descartes' rule of signs, we're going to have 2 or 0 negative roots, and our possible rational roots are:
-10, -10/3, -5, -5/3, -2, -2/3, -1, -1/3

All we have to do is test each one to see if they make the function equal to 0. Once we find one value, we use long division.

p(-10) = 3(-10)^4 + 14(-10)^3 + 23(-10)^2 + 10
= 3(10000) + 14(-1000) + 23(100) + 10
= 30000 - 14000 + 2300 + 10, which is non-zero.

p(-10/3) = 3(10000/81) + 14(-1000/27) + 23(100) + 10
= 10000/27 - 14000/27 + 2300 + 10
= -4000/27 + 2300 + 10
= -4000/27 + 62100/27 + 270/27, which is non-zero.

p(-5) = 3(-5)^4 + 14(-5)^3 + 23(-5)^2 + 10
= 3(625) + 14(-125) + 23(25) + 10
= 1875 - 1750 + 575 + 10
= 125 + 575 + 10, which is non-zero

p(-5/3) = 3(-5/3)^4 + 14(-5/3)^3 + 23(-5/3)^2 + 10
= 3(625/81) + 14(-125/27) + 23(25/9) + 10
= 625/27 - 1750/27 + 1725/27 + 270/27, which is non-zero

p(-2) = 3(-2)^4 + 14(-2)^3 + 23(-2)^2 + 10
= 3(16) + 14(-8) + 23(4) + 10
= 48 - 112 + 92 + 10, which is non-zero.

p(-2/3) = 3(-2/3)^4 + 14(-2/3)^3 + 23(-2/3)^2 + 10
= 3(16/81) + 14(-8/27) + 23(4/9) + 10
= 16/27 - 112/27 + 276/27 + 270/27, which is non-zero

p(-1) = 3(-1)^4 + 14(-1)^3 + 23(-1)^2 + 10
= 3 - 14 + 23 + 10, which is non-zero

p(-1/3) = 3(-1/3)^4 + 14(-1/3)^3 + 23(-1/3)^2 + 10
= 3(1/81) + 14(-1/27) + 23(1/9) + 10
= 1/27 - 14/27 + 69/27 + 270/27, which is non-zero.

Therefore, this isn't factorable as there aren't any rational roots.

Answer 2

Unfactorable unless they want this
x^3+23x^2+10 (3x+14)

Group Factoring... Take out an x^3 in first two... then nothing in the second.

Answer 3

Is that not already done?