Question

find all the zeroes of x3+11x2+23x-35 of two of zeroes are 1and-5

Answer 1

See the picture and plz comment if there is any problem.

Answer 2

1,-5,-7

Step-by-step explanation:

Dividend : $x^3+11x^2+23x-35$

We are given that two of zeroes are 1and-5

So. x=1, x= -5

x-1=0 , x+5-0

So, divisor = $(x-1)(x+5)$

Divisor = $x^2+5x+x^2-5$

$Dividend = (Divisor \times Quotient)+remainder$

$x^3+11x^2+23x-35 = ((x^2+5x+x^2-5) \times x)+(7x^2+28x-35)$

$x^3+11x^2+23x-35 = ((x^2+5x+x^2-5) \times x+7)+0$

Quotient = x+7

So, For another zero = x+7=0

x = -7

Hence the third zero is -7

# Learn more :

Division

1)https://brainly.in/question/3930893, Answered by KishoreJavvadi