Math Practice

Descartes' Rule of Signs Practice Problems

Descartes' rule of signs is two counts and a table. These twelve problems drill the counting in P(x) and P(−x), the 'or fewer by an even number' clause, the no-sign-change case, and the zero-constant-term case where a power of x has to be factored out first.

Questions

1.
How many sign changes are there in ?
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Signs . Changes at and : two. The last pair is not a change.
2.
has 3 sign changes. What are the possible numbers of positive real zeros? Enter them comma separated.
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The count is the number of sign changes or fewer by an even number: or . It cannot be or , because non-real zeros disappear in pairs.
3.
has no sign changes. How many positive real zeros does it have?
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Zero sign changes means zero positive real zeros — there is nothing to reduce by an even number. Every term is positive, so for all , which says the same thing directly.
4.
For , write in standard form.
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Replace with : , , stays as written with a flipped sign. So . Only the odd powers change sign.
5.
For , how many negative real zeros are possible? Enter the possibilities comma separated.
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has signs : one change. One sign change is exact, so there is exactly one negative real zero. (It is ; the others are and .)
6.
. How many sign changes does have?
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Signs : every consecutive pair flips, so five changes. Positive zeros: , or .
7.
. How many negative real zeros does it have?
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: every sign is negative, so no changes and no negative real zeros. Combined with the previous problem, the zeros are , or positive plus , or non-real.
8.
How many sign changes does have? (Remember to skip the missing powers.)
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The non-zero coefficients in order are : two changes. The missing , and terms are skipped — a zero coefficient is never a sign change.
9.
. How many real zeros does it have?
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No sign changes in , so no positive zeros. All powers are even, so — no negative zeros either. The constant means is not a zero. All four zeros are non-real.
10.
A degree- polynomial has 2 sign changes in and 2 in . How many different (positive, negative, non-real) combinations does Descartes' rule allow?
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Positive: or . Negative: or . Every pairing works and leaves an even number of non-real zeros: , , , . Four rows.
11.
has rational-root candidates . After applying Descartes' rule, how many negative candidates can still be zeros?
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, signs : one change, so exactly one negative real zero. At most one of the four negative candidates can work — and it turns out to be , with the positive zeros and .
12.
Apply Descartes' rule of signs to . How many positive real zeros does it have?
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The constant term is zero, so factor: . Apply the rule to : one sign change, so exactly one positive zero (it is ). The factor contributes three times, which is neither positive nor negative, and also gives the one negative zero .

Written by

Zohaib

MPhil in Mathematics, Founder of Mathovia

Published

21 September 2026

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Descartes’ Rule of Signs Practice Problems

The Descartes’ rule of signs lesson reduces to two counts and a table. These twelve problems practise each step, including the cases the rule handles awkwardly — missing terms, no sign changes, and a zero constant term.

The rule

A count of or is exact; anything larger is a ceiling that can only drop in steps of two.

Forming P(−x)

Odd powers flip; even powers and the constant do not.

The table

For degree , every row satisfies

Zero constant term

Factor out the power of , apply the rule to , and count the zeros at the origin separately.

Frequently Asked Questions

What does Descartes' rule of signs say?+

The number of positive real zeros of a polynomial equals the number of sign changes between consecutive coefficients of P(x), or is less than that by an even number. Replacing x with −x and counting again gives the same information for negative zeros.

Why 'or fewer by an even number'?+

Non-real zeros of a real polynomial come in conjugate pairs, so if some of the would-be positive zeros are actually non-real, they go missing two at a time.

How do I find P(−x)?+

Replace every x with −x. Terms with odd powers change sign; terms with even powers and the constant stay the same. Then count the sign changes in the result.

What if there are no sign changes?+

Then there are no zeros of that kind. No changes in P(x) means no positive real zeros; no changes in P(−x) means no negative ones. If both hold and the constant is non-zero, the polynomial has no real zeros at all.

What if the constant term is zero?+

Then x = 0 is a zero, which is neither positive nor negative. Factor out the highest power of x, apply the rule to the remaining factor, and add the zeros at the origin back at the end.

Does the rule find the zeros?+

No — it only counts them. Use it to trim the rational root theorem's candidate list, then find the actual zeros with synthetic division.

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