Question

Show that p(x)=x^3-3x^2+2x-6 has only 1 real zero. Pls guys help me out

Answer 1

According to the Rational Zeros Theorem,

• The possible roots will be ±(factors of 6).
• →Those are between ±6, ±3, ±2, ±1.

If we substitute x=3

→ $\sf{p(3)=27-27+6-6=0}$

Therefore x-3 is one of its factors.

If we divide p(x) by x-3, we will get another factor.

→ $\sf{p(x)=(x-3)(x^2+2)}$

The second factor has zeros at $\sf{x^2+2=0}$.

But it doesn't have any real zero.

∴There is only one real zero x=3.

For your information,

as I've actually divided above, I will show another method here.

If you noticed there are common factors, you can approach with splitting the terms method.

$\sf{p(x)=(x^3-3x^2)+(2x-6)$

$\sf{p(x)=x^2(x-3)+2(x-3)}$

$\sf{p(x)=(x-3)(x^2+2)}$

Answer 2

Step-by-step explanation:

According to the Rational Zeros Theorem,

The possible roots will be ±(factors of 6).

→Those are between ±6, ±3, ±2, ±1.

If we substitute x=3

→ \sf{p(3)=27-27+6-6=0}p(3)=27−27+6−6=0

Therefore x-3 is one of its factors.

If we divide p(x) by x-3, we will get another factor.

→ \sf{p(x)=(x-3)(x^2+2)}p(x)=(x−3)(x

2

+2)

The second factor has zeros at \sf{x^2+2=0}x

2

+2=0 .

But it doesn't have any real zero.

∴There is only one real zero x=3.

For your information,

as I've actually divided above, I will show another method here.

If you noticed there are common factors, you can approach with splitting the terms method.

\sf{p(x)=(x^3-3x^2)+(2x-6)

\sf{p(x)=x^2(x-3)+2(x-3)}p(x)=x

2

(x−3)+2(x−3)

\sf{p(x)=(x-3)(x^2+2)}p(x)=(x−3)(x

2

+2)