Question

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

Answer 1

Answer:

$\mathfrak{\large{ \green{\underline{\underline{Answer}}}}} \\ \large \green{x = 4} \\ \large \green{x = - 5}$

Step-by-step explanation:

$\mathfrak{\large{ \green{\underline{\underline{given}}}}} \large \green{ = 3 \: and \: - 3 \: are \: the \: two \: zeroes \: } \\ \large \green{ \underline{ \underline{polynomial}}} \large \green{ = {x}^{4} + {x}^{3} + 11 {x}^{2} - 9x + 18 } \\ \large \green{x = 3 \: and \: x = - 3} \\ \large \green{(x - 3)(x + 3)} \\ \large \blue{ \underline{ \underline{formula \: of \: ( {x}^{2} - {y}^{2}) = (x + y)(x - y) }}} \\ \large \blue{ {x}^{2} - 9 } \\ \large \blue{ \frac{ {x}^{4} + {x}^{3} + 11 {x}^{2} - 9x + 18 }{ {x}^{2} - 9 } } \\ \large \green{ \underline{ \underline{on \: dividing \: we \: get}}} \\ \large \green{quotient = {x}^{2} + x + 20 } \\ \large \green{remainder = 198} \\ \large \blue{ \underline{ \underline{factorise \: the \: quotient \: into \: middle \: term \: split}}} \\ \large \blue{ {x}^{2} + x + 20 = 0 } \\ \large \blue{ {x}^{2} + 5x - 4x + 20 = 0 } \\ \large \blue{x(x + 5) - 4(x + 5) = 0} \\ \large \blue{(x - 4)(x + 5) = 0} \\ \large \blue{x = 4 \: and \: x = - 5}$

Answer 2

CORRECT QUESTION:

If 3and -3 are the two zeroes of the polynomial (x^4+x³-11x²-9x+18) find all the zeroes of the given polynomial.

ANSWER:

1 and -2 are the other two zeroes of the polynimial

GIVEN:

x=3,x=-3

TO FIND:

ALL ZEROES OF THE GIVEN POLYNOMIAL

EXPLANATION:

(x-3)(x+3)=x²-9 [IDENTITY : (a+b)(a-b)=a²-b²]

Dividing the given equation by x²-9 we get quotiet as x²+x-2

Split the middle term as x²+2x-x-2

x(x+2)-1(x+2)=0

(x-1)(x+2)=0

x=1,x=-2

1 and -2 are the other two zeroes of the polynimial

NOTE: THE CALCULATION IS GIVEN IN THE ATTACHMENT.