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Answer 1

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1. >>>>>>>x3 - 3x2 - 9x - 5

Ans) p(x) = x3 - 3x2 - 9x - 5

Factors of 5 = 1,5,-1,-5

p(1) = 13 - 3 x (1)2  - 9 x (1) - 5

       = 1 -3 - 9 - 5 

       = 1 - 17 = - 16 ≠ 0

p(-1) = (-1)3 x 3 x (-1)2 - 9 x (-1) -5

        = -1 - 3 + 9 - 5 = -9 + 9 = 0

p(-1) =0

∴ (x+1) is a factor of p(x)

    p(x) ÷ (x+1)

    x3 - 3x2 - 9x - 5 ÷ x + 1
= x2 - 4x - 5

x2 - 4x - 5

x2 - 5x + 1x - 5 

= x ( x - 5) + 1 ( x - 5)

= ( x - 5) ( x + 1) Ans

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2. >>>>>>>>>>x^3 + 13x^2 +32x + 20

Let p (x) = x^3 + 13x^2 +32x + 20
Using hit and trial method
x= -1
p (-1) = -1 + 13 -32 +20
=-33+ 33
= 0
So, (x +1 ) is a factor of p (x) [by factor theorem]
To fing other factors we must divide p (x) by (x+1)
And the other factor found is x^2 +12x + 10
Factorising x^2 +12x +20
= x^2 +10x + 2x +20
= x (x +10) +2 (x +10)
= (x+2)(x+10)
So, polynomial p (x) would be factorised as (x+2)(x +10)(x+1). Ans.

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3.>>>>>>>> x3- 23x​2+142x-120
let p(x)=​ x3- 23x​2+142x-120
g(x)=x-1
     0=x-1
     x=1
p(1)=(1)3-23(1)2+142(1)-120
       =1-23+142-120
      =143-143
     =0
Hence x-1 is a factor of ​x3- 23x​2+142x-120
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by long division
 
x3- 23x​2+142x-120 = (x-1)(x2+22x+120)
                                =(x-1)(x2+10x+12x+120)
                                =(x-1)(x(x+10) 12(x+10))
                                =(x-1)(x+12)(x+10) Ans.


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4. >>>>>>> x³-7x+6

(x-1)is the factor of x³-7x+6

so if we divide x³-7x+6 with (x-1) then we will get x²+x-6

so we can write x³-7x+6 as (x-1)(x²+x-6)

=(x-1)(x²+3x-2x-6)

=(x-1)(x(x+3)-2(x+3))
=(x-1)(x-2)(x+3)
so the other zeros are 2 and -3
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5. >>>>>>>>> 2y3-5y2-19y+42

Solution :- See the attachment

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