Finding Other Zeros Using Factor Theorem

We are given the polynomial $P(x) = x^3 - 6x^2 + 11x - 6$ and one of its zeros, $x=2$.

According to the Factor Theorem, if $x=c$ is a zero of a polynomial, then $(x-c)$ is a factor of that polynomial. Since $x=2$ is a zero, $(x-2)$ must be a factor of $P(x)$.

Polynomial Division

To find the other factors, we divide the polynomial $P(x)$ by $(x-2)$. We can use synthetic division:

The result of the division gives the coefficients of the quotient polynomial. The remainder is 0, confirming that $(x-2)$ is a factor. The quotient is a quadratic polynomial: $x^2 - 4x + 3$.

Solving the Quadratic Factor

The remaining zeros of the original polynomial are the roots of the quadratic equation obtained from the quotient: $x^2 - 4x + 3 = 0$.

We can solve this quadratic equation by factoring:

$(x-1)(x-3) = 0$

Setting each factor to zero gives the roots:

• $x-1 = 0 \implies x = 1$
• $x-3 = 0 \implies x = 3$

Therefore, the other two zeros of the polynomial $x^3 - 6x^2 + 11x - 6$ are 1 and 3.