Question

how many zeros dose the polynomial (x-3)² -4 can have? also, find it's zeros​

Answer 1

Answer:

1 and 5

Explanation:

${(x - 3)}^{2} - 4 \\ {x}^{2} - 6x + 9 - 4 \\ {x}^{2} - 6x + 5$

Since, it is a quadratic polynomial it will have two zeroes.

Lets find the zeroes by quadratic equation.

$x = \frac{ - b± \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ -( - 6)± \sqrt{ {( - 6)}^{2} - 4(1)(5)} }{2(1)} \\ \\ x = \frac{6± \sqrt{36 - 20} }{2} \\ \\ x = \frac{6± \sqrt{16} }{2} \\ \\ x = \frac{6±4}{2} \\ \\ x = 5 \: or \: x = 1$

Hence,the zeroes of the polynomial is 5 and 1.

Answer 2

Concept:

The values of x that fulfil the equation f(x) = 0 are the zeros of a polynomial.

Given:

polynomial $(x-3)^{2}-4$

To find:

zeroes of the  polynomial $(x-3)^{2}-4$

Solution:

Expanding the polynomial,

$(x-3)^{2} -4\\x^{2} + 9 - 6x - 4\\x^{2} -6x+ 5$

Since, the polynomial has the the highest degree as 2, it will have two zeroes.

The equation will be

$x^{2} -6x + 5=0$

Splitting the middle term, we have

$x^{2} -(1+5)x + 5=0\\x^{2} -x- 5x+5=0\\x(x-1)-5(x-1)=0$

$(x-1)(x-5)=0$

The zeroes will be

$x-1=0$                                             $x-5=0$

$x=1$                                                    $x=5$

The zeroes of the polynomial $(x-3)^{2}-4$ are $1$ and $5$.

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