Question

If 3 and -3 are two zeros of the polynomial (x4+x3-11x2-9x+18) ,find all the zeros of the given polynomial.

Answer 1

So
A.T.Q
x-3 and x+3
so to divide x4+x3-11x2-9x+18 we have to divide both the zeroes.
(x-3)(x+3)=x²-9(since (a-b)(a+b)=a²-b²)
so I have divide and I found that.
x²+x-2
x²+2x-x-2
x(x+2)-1(x+2)
(x-1)(x+2)
x= 1 ;x= (-2)
the zeroes of x⁴+x³-11x²-9x+18 is 3,(-3),(-2) and 1.
I hope you would like my answer and could be brainiest one.

Answer 2

Step-by-step explanation:

$let \: p(x) = {x}^{4} + {x}^{3} - 11 {x}^{2} - 9x + 18 = 0$

$zeros \: are \: 3 \: and \: - 3$

$therefore \: (x - 3) \: and \: (x + 3) \: are \: its \: factors$

$(x - 3)(x + 3) = {x}^{2} - 9 \: is \: also \: a \: factor$

$let \: g(x) = {x}^{2} - 9$

$dividing \: p(x) \: by \: g(x)$

For dividation refer to the attachment here..

$let \: q(x) = {x}^{2} + x - 2$

${x}^{2} + x - 2 = 0 \\ {x}^{2} + 2x - x - 2 = 0 \\ x(x + 2) - 1(x + 2) = 0 \\ (x + 2)(x - 2) = 0 \\ either \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: or \\ x + 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x - 1 = 0 \\ x = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 1$

$therefore \: all \: the \: zeros \: are \: \\ 3 \\ - 3 \\ 1 \\ - 2$

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