Help with polynomials, please
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Help with polynomials, please

[From: ] [author: ] [Date: 13-05-04] [Hit: ]
Its been awhile since Ive done math, and trying to help my child with homework.Im not sure what 1 is a zero means :-X.Can someone please walk me through this problem?-okay because f(x) = x^3 - 5x^2 +6x -2 has a zero (a zero is when the function touches the y -axis) at 1 do synthetic divisionWith synthetic division you place the coefficient of each degree as I did below and to the left place the zero (in this case the zero is 1) Bring the 1 down as I did and multiply it by the zero (which is 1) and then place the number below the subsequent number. Repeat the same pattern as I did below.......
For the polynomial below, 1 is a zero.

f(x) = x^3 - 5x^2 +6x -2

Express as a product of linear factors.

It's been awhile since I've done math, and trying to help my child with homework. I'm not sure what "1 is a zero" means :-X. Can someone please walk me through this problem?

-
okay because f(x) = x^3 - 5x^2 +6x -2 has a zero (a zero is when the function touches the y -axis) at 1 do synthetic division

With synthetic division you place the coefficient of each degree as I did below and to the left place the zero (in this case the zero is 1)
Bring the 1 down as I did and multiply it by the zero (which is 1) and then place the number below the subsequent number. Repeat the same pattern as I did below.

1| 1 -5 6 -2
|___1_ -4_2__
_ 1 -4 2 0

the 1 -4 2 0 tells us that 1 was indeed a zero and the 1 -4 and 2 tell us the coefficient of each degree

so (1)x^2 + (-4)x + 2
to find zeroes set it equal to 0
do the quadratic formula

x = -b plus or minus the squareroot of b squared minus 4ac all over two times a

a = 1
b=-4
c=2

you will get x = 2 plus or minus radical 2

so (x-1)(x-(2+radical 2))(x-(2-radical 2))

or leaves as (x-1)(x^2 -4x + 2) | if you do not need linear factors

The process is really not as convoluted as some may think.

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What "1 is a zero" means is: If you plug 1 into that (letting x=1 and evaluating it), you get zero as an answer. (1)^3 - 5*(1)^2 + 6*(1) - 2 = 0

A zero of a polynomial is also called a root.

If 'a' is a zero, or root, then (x-a) is a linear factor, so in this case (x-1) is a factor.

You can factor out (x-1) from the cubic polynomial to arrive at a quadratic polynomial.
x^3 - 5x^2 +6x -2 = (x-1)(ax^2 + bx + c)
See, now we can write the cubic as a product of factors. One of those factors is a linear factor and the other is a quadratic factor.
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