Apply Descartes' Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. $$ p(x)=5 x^{6}-7 x^{5}+7 x^{4}+4 x^{3}+6 x^{2}+8 x+2 $$ Each part may have more than one possibility. Use commas to separate multiple responses. Part 1 of 2 Number of possible positive real zeros: Part 2 of 2

Question

UpStudy AI · Accepted Answer

**Step 1: Determine the number of possible positive real zeros**

Consider the polynomial  
$$
p(x)=5 x^{6}-7 x^{5}+7 x^{4}+4 x^{3}+6 x^{2}+8 x+2.
$$  
Examine the coefficients in order:  
$$
+5,\ -7,\ +7,\ +4,\ +6,\ +8,\ +2.
$$  
Count the sign changes:  
- $$+5$$ to $$-7$$ is one sign change.  
- $$-7$$ to $$+7$$ is a second sign change.  
- All remaining coefficients are positive, so there are no additional sign changes.

Thus, there are 2 sign changes, indicating either 2 or 0 positive real zeros.

---

**Step 2: Determine the number of possible negative real zeros**

Replace $$ x $$ with $$-x$$ in the polynomial:  
$$
p(-x)=5(-x)^{6}-7(-x)^{5}+7(-x)^{4}+4(-x)^{3}+6(-x)^{2}+8(-x)+2.
$$  
Simplify each term:  
- $$(-x)^{6}=x^{6}$$ so $$5(-x)^{6}=5x^{6}$$.  
- $$(-x)^{5}=-x^{5}$$ so $$-7(-x)^{5}=-7(-x^{5})=+7x^{5}$$.  
- $$(-x)^{4}=x^{4}$$ so $$7(-x)^{4}=7x^{4}$$.  
- $$(-x)^{3}=-x^{3}$$ so $$4(-x)^{3}=4(-x^{3})=-4x^{3}$$.  
- $$(-x)^{2}=x^{2}$$ so $$6(-x)^{2}=6x^{2}$$.  
- $$8(-x)=-8x$$.  
- The constant term remains $$2$$.

Thus, the polynomial becomes:  
$$
5x^{6}+7x^{5}+7x^{4}-4x^{3}+6x^{2}-8x+2.
$$  
Now, list the signs of the coefficients:  
$$
+,\ +,\ +,\ -,\ +,\ -,\ +.
$$  
Count the sign changes:  
- From $$+$$ to $$+$$ (first to second): no change.  
- From $$+$$ to $$+$$ (second to third): no change.  
- From $$+$$ to $$-$$: one change.  
- From $$-$$ to $$+$$: second change.  
- From $$+$$ to $$-$$: third change.  
- From $$-$$ to $$+$$: fourth change.

There are 4 sign changes, indicating either 4, 2, or 0 negative real zeros.

---

**Final Answers:**

Positive real zeros: $$2,0$$  
Negative real zeros: $$4,2,0$$
Positive real zeros: 2 or 0 Negative real zeros: 4, 2, or 0

Apply Descartes’ Rule of Signs.
Determine the number of possible positive and negative real zeros for the given function.

Each part may have more than one possibility. Use commas to separate multiple responses.
Part 1 of 2
Number of possible positive real zeros:
Part 2 of 2

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Pre Algebra Questions

Apply Descartes’ Rule of Signs. Determine the number of possible positive and negative real zeros for the given function. Each part may have more than one possibility. Use commas to separate multiple responses. Part 1 of 2 Number of possible positive real zeros: Part 2 of 2