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?
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?
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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.
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.
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.