Read polynomial zeros and multiplicities from factored form

Problem

Answer

Step-by-step solution

1. Find each zero

Set each linear factor to zero: $x-5=0$ gives 5, $x-13=0$ gives 13, $x+12=0$ gives -12, and $x+9=0$ gives -9.

2. Read the exponents

The first two factors each occur once, so 5 and 13 have multiplicity one. The factor $(x+12{)}^2$ makes -12 a double zero; $(x+9{)}^3$ makes -9 a triple zero. No expansion is needed.

More examples: crossings, touches and degree

These additional problems were solved on 6 October 2026.

Example 1. Find the real zeros, their multiplicities, the x-axis behavior and the degree-based upper bound on turning points for

The zeros are $x=-1/4$, of multiplicity 2, and $x=-3$, of multiplicity 3. The larger zero is $-1/4$, not $-3$. An even multiplicity gives a touch without a sign change; an odd multiplicity gives a crossing. Thus the graph touches at $-1/4$ and crosses at $-3$. The degree is $2+3=5$. Any degree-5 polynomial has at most $5-1=4$ turning points. This is a degree-based bound, not a claim that this particular polynomial has four turns.

Example 2. Answer the same questions for

For every real $x$, both $3x^2+8$ and $x^2+5$ are strictly positive. Therefore $g(x)>0$: there are no real zeros, and the graph neither crosses nor touches the x-axis. Its degree is $2\cdot 2+2=6$. A degree-6 polynomial has at most $6-1=5$ turning points. Again, this upper bound is not the actual count for the given function.

Why multiplicity controls a crossing

Near a real zero $r$, write the polynomial as $(x-r{)}^{m}h(x)$, where $h(r)\ne 0$. The factor $h(x)$ keeps the same sign sufficiently close to $r$. If $m$ is odd, $(x-r{)}^m$ changes sign across $r$; if $m$ is even, it does not. This explains the crossing-versus-touch rule without guessing from a picture.