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  1. Let f(x)=3x³−10x²−27x+10 5,−2 and 1/3​ are the zeroes of the polynomial ( given ) Therefore, f(5)=3(5)³−10(5)²−27(5)+10 =3×125−250−135+10 =0 f(−2)=3(−2)³−10(−2)²−27(−2)+10 −24−40+54+10 =0 f(1/3)=3(1/3)³−10(1/3)²−27(1/)³+10 =1/8−10/9−9+10 =0 Verify relations : General form of cubic equation :ax³+bx²+Read more

    Let f(x)=31027x+10

    5,2 and 1/3 are the zeroes of the polynomial ( given )

    Therefore,
    f(5)=3(510(527(5)+10
    =3×125250135+10
    =0

    f(2)=3(210(227(2)+10
    2440+54+10
    =0

    f(1/3)=3(1/310(1/327(1/+10
    =1/810/99+10
    =0

    Verify relations :
    General form of cubic equation :a+b+cx+d
    now ,
    Consider α=5,β=2  and y=31

    α+β+y=52+1​/3=10/3=b/a

    αβ+βy+αy=5(2)+(2)(1​/3)+(5×1/3)=10+(−2)​/3+5/3=9=c/a

    and αβy=5(2)(1/3)=−10/3=d/a

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  2. General Form of cubic polynomial whose zeroes are a. b and c is : x³−(a+b+c)x²+(ab+bc+ca)x−abc..........(1) To find : cubic polynomial whose zeroes are 2,−3 and 4 Let us say , a=2,b=−3 and c=4 Putting the values of a, b and c in equation (1) we get : =x³−(2−3+4)x²+(−6−12+8)x−(−24) =x³−3x²−10x+24 whiRead more

    General Form of cubic polynomial whose zeroes are a. b and c is :
    (a+b+c)+(ab+bc+ca)xabc..........(1)
    To find : cubic polynomial whose zeroes are 2,3 and 4

    Let us say , a=2,b=3 and c=4
    Putting the values of a, b and c in equation (1) we get :

    =(23+4)+(612+8)x(24)
    =310x+24
    which is required polynomial.

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  3. General Form of cubic polynomial whose zeroes are a. b and c is : x³−(a+b+c)x²+(ab+bc+ca)x−abc..........(1) To find : cubic polynomial whose zeroes are 1/2​,1 and -3 Let us say , a=1/2​,b=1 and c=−3 Putting the values of a,b and c in equation (1) we get : =x³−(1/2+1−3)x²+(1/2​−3−3/2​)x−(−3/2​) =x³−(Read more

    General Form of cubic polynomial whose zeroes are a. b and c is :
    (a+b+c)+(ab+bc+ca)xabc..........(1)

    To find : cubic polynomial whose zeroes are 1/2,1 and –3

    Let us say , a=1/2,b=1 and c=3
    Putting the values of a,b and c in equation (1) we get :

    =(1/2+13)+(1/23−3/2)x(−3/2)

    =(−3)/24x+3/2

    =2+38x+3
    which is required polynomial.

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  4. This answer was edited.

    As we know , general form of cubic polynomial whose zeroes are a, b and c is : x³−(a+b+c)x²+(ab+bc+ca)x−abc also written as : x³−(sum of zeroes)x²+(sum of the product of zeroes taking two at a time)x−(product of zeroes).............(1) Given : Sum of zeros =5 Sum of the product of its zeros taken twRead more

    As we know , general form of cubic polynomial whose zeroes are a, b and c is :
    x³−(a+b+c)x²+(ab+bc+ca)x−abc

    also written as :
    x³−(sum of zeroes)x²+(sum of the product of zeroes taking two at a time)x−(product of zeroes)………….(1)

    Given :
    Sum of zeros =5
    Sum of the product of its zeros taken two at a time=−2
    Product of its zeros =−24
    Putting these values in equation (1) we get:
    x³−5x²−2x+24
    which is required polynomial.

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  5. Let f(x)=x³−2x²−5x+6 3,−2 and 1 are the zeroes of the polynomial ( given ) Therefore, f(3)=(3)³−2(3)²−5(−2)+6 =27−18−15+6 =0 f(−2)=(−2)³−2(−2)²−5(−2)+6 −8−8+10+6 =0 f(1)=(1)³−2(1)²−5(1)+6 =1−2−5+6 =0 Verify relations : General form of cubic equation :ax³+bx²+cx+d now , Consider α=3,β=−2 and y=1 α+β+Read more

    Let f(x)=25x+6
    3,2 and 1 are the zeroes of the polynomial ( given )

    Therefore,
    f(3)=(32(35(2)+6
    =271815+6
    =0

    f(2)=(22(25(2)+6
    88+10+6
    =0

    f(1)=(12(15(1)+6
    =125+6
    =0

    Verify relations :
    General form of cubic equation :a+b+cx+d
    now ,
    Consider α=3,β=2 and y=1
    α+β+y=32+1=2=b/a
    αβ+βy+αy=3(2)+(2)(1)+1(3)=5=c/a
    and αβy=3(2)(1)=6=d/a

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  6. Given that. AB II CD In Δ AOB and Δ COD ∠OAB = ∠OCD (Alternate interior angles) ∠OBA = ∠ODC (Alternate interior angles) ∠AOB = ∠DOC (Vertically opposite angles) ΔAOB ~ ΔCOD (AAA similarly) We know that, Areabof ΔAOB/Area of ΔCOD ​(AB/CD​)2=(AO/CO​)2=(OB/OD​)2 (2CD/CD​)2 = 4:1

    Given that.
    AB II CD
    In Δ AOB and Δ COD
    OAB = OCD (Alternate interior angles)
    OBA = ODC (Alternate interior angles)
    AOB = DOC (Vertically opposite angles)
    ΔAOB ~ ΔCOD (AAA similarly)
    We know that,
    Areabof ΔAOB/Area of ΔCOD (AB/CD)2=(AO/CO)2=(OB/OD)2
    (2CD/CD)2 = 4:1

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