Question

One zero of a polynomial \[3{x^3} + 16{x^2} + 15x - 18\] is\[\dfrac{2}{3}\]. Find other zeros of polynomials.

Answer 1

Hint: Here we use the given value of zero of the polynomial and use the concept of zeroes of a polynomial that they give polynomial value equal to zero when substituted in place of variable.
* A polynomial of degree \[n\] is a polynomial where the variable has highest power n. It can be written as \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\]
* Any polynomial can have repetitive zeroes, i.e. the zeroes of a polynomial need not be different from each other.
* If \[x = a\] is one of the zeroes of a polynomial \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\] then we divide the polynomial by \[x - a\] to find other zeros.
* Long division method: when dividing\[a{x^n} + b{x^{n - 1}} + ....c\] by\[px + q\] we perform as
\[px + q)\overline {a{x^n} + b{x^{n - 1}} + ....c} ((a/p){x^{n - 1}} + ...\]
            \[\underline { - a{x^n} + (qa/p){x^{n - 1}}} \]
                          \[0.{x^n} + (b - qa/p){x^{n - 1}}\]

Complete step-by-step answer:
We are given one zero of the polynomial\[3{x^3} + 16{x^2} + 15x - 18\] is\[\dfrac{2}{3}\], So we can write \[x = \dfrac{2}{3}\] when substituted in the polynomial gives the value zero.
So, to find other zeros of the polynomial we divide the polynomial by \[x - \dfrac{2}{3}\]
Using long division method:
\[x - \dfrac{2}{3})\overline {3{x^3} + 16{x^2} + 15x - 18} (3{x^2} + 18x + 27\]
          \[\underline {3{x^3} - 2{x^2}} \]
          \[18{x^2} + 15x\]
           \[\underline {18{x^2} - 12x} \]
                     \[27x - 18\]
                     \[\underline {27x - 18} \]
                              \[0\]
Therefore, from long division method we write
\[(x - \dfrac{2}{3}).(3{x^2} + 18x + 27) = 3{x^3} + 16{x^2} + 15x - 18\]
Now we solve for zeros of a quadratic equation \[3{x^2} + 18x + 27\].
Write \[3{x^2} + 18x + 27 = 0\] and taking three common we can make the equation much simpler.
\[3({x^2} + 6x + 9) = 0\]
Dividing both sides of the equation by three
\[{x^2} + 6x + 9 = 0\]
We know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
If we take \[a = x,b = 3\]
Then using the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
\[{(x + 3)^2} = {x^2} + {3^2} + 2 \times 3 \times x\]
\[{(x + 3)^2} = {x^2} + 9 + 6x\]
So, we can write \[{x^2} + 6x + 9 = 0\] as \[{(x + 3)^2} = 0\]
We can write \[(x + 3) \times (x + 3) = 0\]
Equating both the parts in LHS equal to zero separately, we get
\[ \Rightarrow x + 3 = 0\]
Shift the constant value to one side of the equation.
\[ \Rightarrow x = - 3\]
But this is a repetitive zero.
Therefore, remaining zeros of the polynomial are \[ - 3, - 3\]
Thus when one of zero of a polynomial \[3{x^3} + 16{x^2} + 15x - 18\] is \[\dfrac{2}{3}\] then other zeroes are \[ - 3, - 3\]

Note: Students are likely to make mistakes in the process of long division as they forget to change the sign before the next step. Also, note that a cubic polynomial can have at most three zeroes.