A maths problem regarding zeroes of a polynomial?

i have a problem in maths

i m studying in class tenth

it was given in my NCERT book that

suppose we take

2x^2 - 8x + 6 = 2(x-1)(x-3)

then it was given that

x-1=0

x-3=0

so 1 and 3 are zeroes of the polynomials

my problem is that "how we can say that "x-3=0" and "x-1=0"

it was nowhere mentioned about dis???

plzz help urgently!

Update:

i have a problem in maths

i m studying in class tenth

it was given in my NCERT book that

suppose we take

2x^2 - 8x + 6 = 2(x-1)(x-3)

then it was given that

x-1=0

x-3=0

so 1 and 3 are zeroes of the polynomials

my problem is that "how we can say that "x-3=0" and "x-1=0"

it was nowhere mentioned about dis???

plzz help urgently!

hey i saw jake's answer n i think its best

i could understand a bit

but then until u explained was ok

after that

either of the numbers (x-1, x-3) must be zero

i agree to that but in my book its been explained

that a quadratic polynomial has two zeroes n when it is made into a graph it forms a parabola

so i wanted to ask that how can we both of them i.e. x-1, x-3 as zeroes?????

7 Answers

Relevance
  • Anonymous
    10 years ago
    Favorite Answer

    well you should understand what zero of a polynomial means

    it is what values of x will make the polynomial zero

    now 2(x-1)(x-3) is your polynomial

    zero of the polynomial means 2(x-1)(x-3) = 0 or (x-1)(x-3) = 0

    this is the product of two numbers, lets ignore 2 for the time being

    the product of two numbers x-1 and x-3 is zero

    can you think of two numbers which multiply to get zero ?

    the only way you can get zero by multiplication is when one of the numbers is zero

    which means either x-3 = 0 ( implying x = 3) or x-1 = 0 ( implying x=1)

    i hope this answers and clarifies your question

    its pretty straightforward

    • Login to reply the answers
  • Anonymous
    10 years ago

    For this Particular problem zero's of the Polynomial means the Solution of 2x^2 - 8x + 6 = 0, or

    2(x-1)(x-3)=0, or the value of x for which value of the expression becomes zero. thus if 2(x-1)(x-3)=0 then it means product of 2 , (x-1) & (x-3) zero. Since 2 cant be equl to 0, so either the value of x-1 or that of x-3 = 0. which implies that either x=1 or 3.

    Thats it.

    • Login to reply the answers
  • 10 years ago

    Let's take a general polynomial y=(x-a)(x-b)(x-c)..............(x-n). Any zero will make it zero.

    =>to find a zero we can write a polynomial equation (x-a)(x-b)(x-c)..............(x-n) = 0. Put x = a, then the factor (x-a) will be 0 as (a-a) is zero and no matter if you multiply any quantity it will be 0. Similarly b, c, ........n all make it = 0. Hence in maths we state that as a, b, c, d.......n, make this polynomial = 0 these are its zeroes.

    In your case y = 2(x-1)(x-3).

    It will be zero if x-1 =0 or x-3 = 0. It implies x = 1 and x = 3, both make it = 0.

    So these are it's zeroes.

    Since, equating each factor to 0 will give a zero of polynomial, we writ x-a=0 , x-b=0.....x-n = 0

    • Login to reply the answers
  • 10 years ago

    x-1=0 so what is x? x must be 1, since 1-1=0

    Like wise, x-3=0, then x must be three.

    They equate it to 0 so you can find the value of the x.

    • Login to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • 10 years ago

    2x^2 - 8x + 6 = 0

    So 2(x-1)(x-3) = 0

    So either

    x- 1 = 0 i.e x = 1

    OR

    x- 3 = 0 i.e x = 3

    • Login to reply the answers
  • 10 years ago

    we have the eqn f(x) = 2(x-1)(x-3);

    Zeroes of the polynomials are points which it passes through the x-axis, to get these points, our f(x) should be zero. so in this eqn, x must be 1 or 3 for the eqn to be zero . . . :))

    • Login to reply the answers
  • Anonymous
    10 years ago

    u dont have any other methodd for it

    either one has to be taken as zero otherwise there wud b no number by which if we put in place of x then it will nil the whole eqn

    • Login to reply the answers
Still have questions? Get your answers by asking now.