Studdy Solution
STEP 1
What is this asking?
We're given a polynomial P(x) and told that 5i is one of its zeros.
We need to find all the other zeros and write P(x) as a product of linear factors.
Watch out!
Complex roots always come in conjugate pairs!
Don't forget that if 5i is a root, so is −5i!
STEP 2
1. Find the quadratic factor.
2. Polynomial long division.
3. Factor the remaining cubic.
STEP 3
Since we know 5i is a zero, and complex zeros come in conjugate pairs, we know −5i *must* also be a zero!
This means (x−5i) and (x+5i) are factors of P(x).
STEP 4
Let's **multiply these factors** together to get a quadratic factor:
(x−5i)(x+5i)=x2+5ix−5ix−25i2=x2−25i2
Remember that i2=−1, so we have:
x2−25(−1)=x2+25
So, (x2+25) is a factor of P(x)!
STEP 5
Now, we'll **perform polynomial long division** to divide P(x) by (x2+25).
This will give us another factor of P(x).
3x3x2+253x50−8x40−x3010x20−8x23x5−8x4−x310x20−5x−8x4+75x3−x3−200x2+10x2−25x0+2500+10+74x3−190x2−25x+250−190x2−25x+250
Great! The division worked out perfectly, with no remainder.
This means 3x3−8x2−5x+10 is another factor.
STEP 6
We can **factor the cubic** 3x3−8x2−5x+10 by grouping.
3x3−8x2−5x+10=x2(3x−8)−5(x−2)
Oops, that didn't work!
Let's try factoring by grouping a different way.
3x3−8x2−5x+10=3x3−5x−8x2+10=x(3x2−5)−2(4x2−5)
Hmm, still no luck.
Let's try rational root theorem!
STEP 7
The possible rational roots are ±1,±2,±5,±10,±31,±32,±35,±310.
Let's try x=2:
3(2)3−8(2)2−5(2)+10=24−32−10+10=−8
Nope, not a root.
Let's try x=32:
3(32)3−8(32)2−5(32)+10=98−932−310+10=98−32−30+90=936=4
Still no luck.
Let's try x=35.
3(35)3−8(35)2−5(35)+10=9125−9200−325+10=9125−200−75+90=90=0
Yes! x=35 is a root!
So (3x−5) is a factor.
STEP 8
Now we can do polynomial long division again, dividing 3x3−8x2−5x+10 by (3x−5).
x23x−53x30−3x20−10x0−x3x3−5x2−3x2+5x−10x+100−2−8x2−5x+10−5x+10
So we have x2−x−2=(x−2)(x+1).
STEP 9
P(x)=(x−5i)(x+5i)(3x−5)(x−2)(x+1)
