Question

What is a quartic function with only the two real zeros given?

x = -4 and x = -1

Answer 1

A. y = x4 + 5x3 + 5x2 + 5x + 4
B. y = x4 - 5x3 - 5x2 - 5x - 4
C. y = -x4 + 5x3 + 5x2 + 5x + 4
D. y = x4 + 5x3 + 5x2 + 5x - 5

Answer 2

I believe it's D because -4 and -1 is -5

Answer 3

if (x-a) is a factor of P(x) we would expect:
\[P(x)=(x-a)Q(x)\]
\[P(a)=(\underbrace{a-a}_{=0})Q(a)\]

Answer 4

we can divide out the roots ...

Answer 5

well if x = -4 then x + 4 is a factor

and

x =-1 means x + 1 is a factor so the quadratic is

f(x) = (x + 4)(x + 1)

now just expand it

Answer 6

x4 + 5x3 + 5x2 + 5x - 5
0 -4 -4 -4 -4
-4 1 1 1 1 -9
^^ not 0, -4 aint a root of D

Answer 7

x4 + 5x3 + 5x2 + 5x + 4
0 -4 -4 -4 -4
-4 1 1 1 1 0

1 1 1 1
0 -1 0 -1
-1 1 0 1 0

x^2+1 is whats left after we factor out the other 2 roots.

Answer 8

@amistre64 can you explain what you mean with a lot less numbers please?

Answer 9

prolly not, its take more words than numbers.

follow campbells approach

Answer 10

D doesnt have -4 as a root, so D is out

Answer 11

test it out for yourself

Answer 12

@amistre64 =(x+4)(x+1)
=(x)(x)+(x)(1)+(4)(x)+(4)(1)
=x^2+x+4x+4
=x^2+5x+4???

Answer 13

@amistre64 does that look right?

Answer 14

no, xx is a root of 0

Answer 15

wait, i didnt read it correctly

Answer 16

(x+4)(x+1) = x^2 + 5x + 4

what we want is 2 more facotrs that are not real ...

some (x-(a + bi))(x-(a-bi))

Answer 17

multiplying it all out is going to e a bother, and prolly not that useful.

the division of the real roots i think is more beneficial

Answer 18

now if we look ahead a bit, we would want some other factor to be x^2 + n for some n>0 so that it has no real roots left in it

(x^2 + 5x + 4)(x^2 + n)

x^2(x^2 + n) + 5x(x^2 + n) + 4(x^2 + n)

x^4 +n x^2 + 5x^3 + 5n x + 4x^2 + 4n

x^4 + 5x^3 +n x^2+ 4x^2 +5n x + 4n

x^4 + 5x^3 +(4+n) x^2+ 5n x + 4n

now we can compare this part by part to see what options fit

Answer 19

Start at the last one you wrote and evaluate? @amistre64

Answer 20

yeah, its not D, so the last term has to be either +4 or -4

let n=+1 or -1 and see what matches

Answer 21

lol, we said n cant be less than 0 .. soo n=1 has to be it

Answer 22

Okay. So n =1 I got
x4+5x3+(4+1)(x2)+(5)(1)x+(4)(1)
x4+5x3+(4+1)(x2)+(5)(1)x+(4)(1)
=x4+5x3+5x2+5x+4
@amistre64

Answer 23

thats what we are looking for

Answer 24

@amistre64 I can't believe I figured it out. Thank you so much for all your help. You are a savior

Answer 25

good luck :)