SOLUTION: The equation x⁴-2x²-16x-15 has four roots. If two of the roots are real and equal to x=-1 and x=3, by a process of long division and solving a quadratic equation find the t

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The equation x⁴-2x²-16x-15 has four roots. If two of the roots are real and equal to x=-1 and x=3, by a process of long division and solving a quadratic equation find the t      Log On


   



Question 475872: The equation x⁴-2x²-16x-15 has four roots. If two of the roots are real and equal to x=-1 and x=3, by a process of long division and solving a quadratic equation find the two complex roots
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
given:
+x+=+-1+
+x+%2B+1+=+0+
and
+x+=+3+
+x+-+3+=+0+
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Complex roots come in pairs
of the form +a+%2B+bi+
and +a+-+bi+
I can factor out +%28x+%2B+1%29%2A%28x+-+3%29+
+%28x+%2B+1%29%2A%28x+-+3%29+=+x%5E2+-+2x+-+3+
+%28x%5E4+-+2x%5E2+-+16x+-+15%29+%2F+%28x%5E2+-+2x+-+3%29+=+x%5E2+%2B+2x+%2B+5+
I can use quadratic formula now
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+2+
+c+=+5+
x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A1%2A5+%29%29%2F%282%2A1%29+
x+=+%28-2+%2B-+sqrt%28+4+-+20+%29%29%2F2+
x+=+%28-2+%2B-+sqrt%28+-16+%29%29%2F2+
x+=+%28-2+%2B+4i%29%2F2+
+x+=+-1+%2B+2i+ 1st root
and
x+=+%28-2+-+4i%29%2F2+
+x+=+-1+-+2i+ 2nd root