Question

If 2 zeroes of the polynomial p(x)=x^3+2x^2-9x_18 are 3 and -3 find the other zero of the polynomial p(x)

Answer 1

Answer:

( x+3)(x-3) are the factors of x4+x3-11x2-9x+18

=> ( x+3)(x-3)= x2-32

= x2-9

therefore x2+0x-9 is a factor of x4+x3-11x2-9x+18

therefore dividing x4+x3-11x2-9x+18 by x2+0x-9 =x2+x-2

there fore

x4+x3-11x2-9x+18 = (x2+x-2)(x2-9)

= (x+1)(x-2)(x+3)(x-3)

=> the zeroes of polynomial x4+x3-11x2-9x+18 = -1, 2 , -3, 3

Answer 2

Aɴꜱᴡᴇʀ

$\huge \sf{} - 2 \: is \: the \: other \: zero$

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Gɪᴠᴇɴ

$\large \sf{}p(x) = {x}^{3} + 2 {x}^{2} - 9x - 18$

Two zeros of the polynomia p(x)= 3 and -3

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ᴛᴏ ꜰɪɴᴅ

The other zero?

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Sᴛᴇᴘꜱ

Given that two zeros of the polynomia are 3 and -3 ,then (x-3) and(x+3) are its zeros.

So now lets multiply them both....

$\large\sf{}(x - 3)( x + 3) \\ \\ \large \sf{} x(x + 3) - 3(x+ 3) \\ \\ \large \sf{} \rightarrow {x}^{2} + \cancel{ 3x} - \cancel{} - 3x + 3 \\ \\ \large \sf{} \implies {x}^{2} + 3 \\ \\ \large \sf{}so \: \: \: \: \frac{\large \sf{} {x}^{3} + 2 {x}^{2} - 9x - 18}{ {x }^{2} + 3 } \\ \\ \large \sf{}now \: see \: attachment \\ \\ \large \sf{}thus \: the \: other \: zero \: is \: \huge \bold{} - 2 \: or \: (x + 2)$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$