Find a polynomial P(x) that has degree 5 with integer coefficient and zeros 12,−1 and −i, and leading coefficient 4; the zero −1 has multiplicity 2.
Recall that if the polynomial function P has real coefficient and if a a+bi is a zero of P, then a−bi is also a zero of P. In our case, we have 12,−1,−i and i. Thus the required polynomial has the form.
P(x)=a(x−12)[x−(−1)]2(x−i)[x−(−i)]ModelP(x)=a(x−12)(x+1)2(x−i)(x+i)SimplifyP(x)=a(x−12)(x+1)2(x2−i2)Difference of squaresP(x)=a(x−12)(x+1)2(x2+1)Recall that i2=−1 P(x)=a(x+1)2[x3+x−12x2−12]Apply FOIL methodP(x)=a[x2+2x+1][x3+x−12x2−12]ExpandP(x)=a[x5+x3−12x4−12x2+2x4+2x2−x3−x+x3+x−12x2−12]ExpandP(x)=a[x5+32x4+x3+x2−12]Simplify and combine like termsP(x)=4[x5+32x4+x3+x2−12]Substitute a=4 to be the leading coefficientP(x)=4x5+6x4+4x3+4x2−2
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