Answer to Question #296383 in Algebra for Jen

Question

1. Solve x2 + 2x − 15 = 0 by factoring. Find both the sum and the product of the zero.

2. Suppose the monic (leading coefficient is one) quadratic polynomial x2+ax+b has zeros r1 and

r2. Write the polynomial in factored form. Multiply the factored form out. What relationships

do you find between this product and the coefficients of the original polynomial.

3. Factor the polynomial x3 + 5x2− 2x- 2

4. Find the sum of the zeros and the product of the zeros. Now find the sum of all the double products of the zeros (i.e., r1r2 + r1r3 + r2r3). What is the relationship between these three values and the coefficients of the polynomial?

4. Suppose the monic cubic polynomial x3 + ax2 + bx + c has zeros r1, r2, and r3. Write the polynomial in factored form and multiply the factors. Write the relationships between the coefficients of each form of the polynomial.

5. Suppose the monic quartic polynomial x4 + ax3 + bx2 + cx + d has zeros r1, r2, r3, and r4.

Answer 1

1]

$0=x^2 + 2x − 15 = x^2+5x-3x-15=x(x+5)-3(x+5)=(x-3)(x+5)\Rightarrow x=-5,3$

Sum of the zeroes is: $-5+3=-2$

Product of the zeroes is:$-5\times 3=-15$

2]

$(x-r_1)(x-r_2)=x^2+x(-r_1-r_2)+r_1r_2$

Since $r_1$ and $r_2$ are roots of $x^2+ax+b=0$ then,

$0=x^2+ax+b=x^2+x(-r_1-r_2)+r_1r_2$

Thus, the relationships between this product and the coefficients of the original polynomial are;

$a=-(r_1+r_2)$

$b=r_1r_2$

3]

$x^3 + 5x^2− 2x- 2=(x+5.305897529)(x-0.7856670112)(x+0.4797694819)$

$\Rightarrow r_1=-5.305897529,\ r_2=0.7856670112,\ r_3=-0.4797694819$

Sum of the zeros: $r_1+r_2+r_3=-5$

Product of the zeros: $r_1\times r_2\times r_3=2$

The sum of all the double products of the zeros: $r_1r_2+r_1r_3+r_2r_3=-2$

The relationship between these three values and the coefficients of the polynomial are;

• Coefficient of $x^2=-(r_1+r_2+r_3)=5$
• Constant term or coefficient of $x^1=x=r_1r_2+r_1r_3+r_2r_3=-2$
• Coefficient of $x^0=-r_1r_2r_3=-2$

4]

$(x-r_1)(x-r_2)(x-r_3)=x^3+x^2(-r_1-r_2-r_3)+x(r_1r_2+r_1r_3+r_2r_3)-r_1r_2r_3$

Since $r_1,\ r_2$ and $r_3$ are roots of $x^3+ax^2+bx+c=0$ then,

$0=x^3+ax^2+bx+c=x^3+x^2(-r_1-r_2-r_3)+x(r_1r_2+r_1r_3+r_2r_3)-r_1r_2r_3$

Thus, the relationships between the coefficients of each form of the polynomial are;

$a=-(r_1+r_2+r_3)$

$b=r_1r_2+r_1r_3+r_2r_3$

$c=-r_1r_2r_3$

5]

$(x-r_1)(x-r_2)(x-r_3)(x-r_4)\\ \qquad\qquad=x^4+x^3(-r_1-r_2-r_3-r_4)+x^2(r_1r_2+r_1r_3+r_2r_3+r_1r_4+r_2r_4+r_3r_4)\\ \qquad\qquad\qquad\qquad-x(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)+r_1r_2r_3r_4$

Since $r_1,\ r_2,\ r_3$ and $r_4$ are roots of $x^4 + ax^3 + bx^2 + cx + d$ then,

$0=x^4 + ax^3 + bx^2 + cx + d\\ \qquad\qquad=x^4+x^3(-r_1-r_2-r_3-r_4)+x^2(r_1r_2+r_1r_3+r_2r_3+r_1r_4+r_2r_4+r_3r_4)\\ \qquad\qquad\qquad\qquad-x(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)+r_1r_2r_3r_4$

Thus, the relationships between the coefficients of each form of the polynomial are;

$a=-(r_1+r_2+r_3+r_4)$

$b=r_1r_2+r_1r_3+r_2r_3+r_1r_4+r_2r_4+r_3r_4$

$c=-(r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4)$

$d=r_1r_2r_3r_4$