SOLUTION: Please help me answer this question. Find the real zeros of the polynomial. State the multiplicity of each real zero. p(x) = 25x5 − 95x4 + 139x3 − 97x2 + 32x &#8722

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me answer this question. Find the real zeros of the polynomial. State the multiplicity of each real zero. p(x) = 25x5 − 95x4 + 139x3 − 97x2 + 32x &#8722      Log On


   

Question 997279: Please help me answer this question.
Find the real zeros of the polynomial. State the multiplicity of each real zero.
p(x) = 25x5 − 95x4 + 139x3 − 97x2 + 32x − 4
Thank you!
Found 2 solutions by fcabanski, MathLover1:
Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Graph to find a zero. Then use synthetic division to factor out that zero. Alternatively, you can use the factors of the constant divided by the factors of the leading coefficient (+ or -) to find potential zeroes. Then check them until you find a zero (no remainder with synthetic division).
It's easier to find zeroes by graphing.
.

Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
p%28x%29+=+25x%5E5+-95x%5E4+%2B+139x%5E3+-97x%5E2+%2B+32x+-+4+...to factor it, first write -95x%5E4 as -20x%5E4-75x%5E4, 139x%5E3 as 4x%5E3%2B+60x%5E3%2B75x%5E3, -97x%5E2 as -60x%5E2-25x%5E2, and 32x as 12x%2B20x, then group



p%28x%29+=%28x%5E3-3x%5E2%2B3x-1%29+%2825x%5E2-20x%2B4%29
p%28x%29+=%28x-1%29%5E3+%285x-2%29%5E2

so, zeros are:
if 0=%28x-1%29%5E3+=>x=1-the multiplicity of this real zero is 3
if 0=%285x-2%29%5E2=>5x-2=0=>x=2%2F5-the multiplicity of this real zero is 2
+graph%28+600%2C+600%2C+-4%2C+4%2C+-5%2C+5%2C+%28x-1%29%5E3+%285x-2%29%5E2%29+