Exercise 13 Page 965

We want to find the number of distinct real roots for the given equation. We will start by factoring out the greatest common factor, x. x^4 +3x^3-4x = 0 ⇕ x(x^3+3x^2-4) = 0 Now, let's recall the Zero-Product Property. If ab=0, then a=0 or b=0This means that we can divide the given equation into two polynomials. x = 0 x^3+3x^2-4= 0 The first polynomial indicates that 0 is a root of the given equation. The remaining roots can be found by solving the second polynomial. To do it, let's start by rewriting the squared term as the sum of two terms so that it is easier to factor by grouping.

x^3+3x^2-4=0

WriteSum

Write as a sum

x^3+2x^2+x^2-4=0

FactorOut

Factor out x^2

x^2(2+x)+x^2-4=0

CommutativePropAdd

Commutative Property of Addition

x^2(x+2)+x^2-4=0

Rewrite

Rewrite 4 as 2^2

x^2(x+2)+x^2-2^2=0

FacDiffSquares

a^2-b^2=(a+b)(a-b)

x^2(x+2)+(x+2)(x-2)=0

Notice that now we can factor out (x+2). Then, we will rewrite the linear term as a difference of two terms to continue factoring.

x^2(x+2)+(x+2)(x-2)=0

FactorOut

Factor out (x+2)

(x^2+x-2)(x+2)=0

WriteDiff

Write as a difference

(x^2+2x-x-2)(x+2)=0

▼

Factor

FactorOut

Factor out x

(x(x+2)-x-2 )(x+2)=0

FactorOutNegSign

Factor out a minus sign

(x(x+2)-(x+2))(x+2)=0

FactorOut

Factor out (x+2)

((x-1)(x+2))(x+2)=0

RemovePar

Remove parentheses

(x-1)(x+2)(x+2)=0

Finally, we will apply the Zero-Product Property to find the roots.

(x-1)(x+2)(x+2)=0

ZeroPropMult

Zero Property of Multiplication

lcx-1=0 & (I) x+2=0 & (II) x+2=0 & (III)

▼

Solve for x

AddEqn

(I):LHS+1=RHS+1

lx=1 x+2=0 x+2=0

(II), (III): LHS-2=RHS-2

lx=1 x=- 2 x=- 2

Now that we have the roots for both polynomials, we can write all the real roots of the original equation. x_1 = 0, x_2= 1 , x_3=- 2 , x_4=- 2 Note that - 2 is a double root, so the given equation has only three distinct real roots. This corresponds to option C.