1]
0=x2+2x−15=x2+5x−3x−15=x(x+5)−3(x+5)=(x−3)(x+5)⇒x=−5,3
Sum of the zeroes is: −5+3=−2
Product of the zeroes is:−5×3=−15
2]
(x−r1)(x−r2)=x2+x(−r1−r2)+r1r2
Since r1 and r2 are roots of x2+ax+b=0 then,
0=x2+ax+b=x2+x(−r1−r2)+r1r2
Thus, the relationships between this product and the coefficients of the original polynomial are;
a=−(r1+r2)
b=r1r2
3]
x3+5x2−2x−2=(x+5.305897529)(x−0.7856670112)(x+0.4797694819)
⇒r1=−5.305897529, r2=0.7856670112, r3=−0.4797694819
Sum of the zeros: r1+r2+r3=−5
Product of the zeros: r1×r2×r3=2
The sum of all the double products of the zeros: r1r2+r1r3+r2r3=−2
The relationship between these three values and the coefficients of the polynomial are;
- Coefficient of x2=−(r1+r2+r3)=5
- Constant term or coefficient of x1=x=r1r2+r1r3+r2r3=−2
- Coefficient of x0=−r1r2r3=−2
4]
(x−r1)(x−r2)(x−r3)=x3+x2(−r1−r2−r3)+x(r1r2+r1r3+r2r3)−r1r2r3
Since r1, r2 and r3 are roots of x3+ax2+bx+c=0 then,
0=x3+ax2+bx+c=x3+x2(−r1−r2−r3)+x(r1r2+r1r3+r2r3)−r1r2r3
Thus, the relationships between the coefficients of each form of the polynomial are;
a=−(r1+r2+r3)
b=r1r2+r1r3+r2r3
c=−r1r2r3
5]
(x−r1)(x−r2)(x−r3)(x−r4)=x4+x3(−r1−r2−r3−r4)+x2(r1r2+r1r3+r2r3+r1r4+r2r4+r3r4)−x(r1r2r3+r1r2r4+r1r3r4+r2r3r4)+r1r2r3r4
Since r1, r2, r3 and r4 are roots of x4+ax3+bx2+cx+d then,
0=x4+ax3+bx2+cx+d=x4+x3(−r1−r2−r3−r4)+x2(r1r2+r1r3+r2r3+r1r4+r2r4+r3r4)−x(r1r2r3+r1r2r4+r1r3r4+r2r3r4)+r1r2r3r4
Thus, the relationships between the coefficients of each form of the polynomial are;
a=−(r1+r2+r3+r4)
b=r1r2+r1r3+r2r3+r1r4+r2r4+r3r4
c=−(r1r2r3+r1r2r4+r1r3r4+r2r3r4)
d=r1r2r3r4
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