Polynomial Factor Theorem Pdf

Or: how to avoid Polynomial Long Division when finding factors

  1. Polynomial Factor Theorem Worksheet
  2. Polynomial Remainder Theorem Calculator
  3. Polynomial Factor Theorem Pdf Download
  4. Polynomial Factor Theorem Pdf
  5. Division Theorem For Polynomials
Pdf

Do you remember doing division in Arithmetic?

'7 divided by 2 equals 3 with a remainder of 1'

CHAPTER 18 POLYNOMIAL DIVISION AND THE FACTOR AND REMAINDER THEOREMS. EXERCISE 74 Page 161. 2) by (x + y) 2. Use the factor theorem to factorize: x. Then a factor (x – a) would exist. State the possible rational zeros for each function. 1) f (x) = 3x2 + 2x − 1 2) f (x) = x6 − 64 3) f (x) = x2 + 8x + 10 4) f (x) = 5x3 − 2x2 + 20 x − 8 5) f (x) = 4x5 − 2x4 + 30 x3 − 15 x2 + 50 x − 25 6) f (x) = 5x4 + 32 x2 − 21 7) f (x) = x3 − 27 8) f (x) = 2x4 − 9x2 + 7. State the possible rational zeros for each function.

Each part of the division has names:

Which can be rewritten as a sum like this:

Factor each polynomial completely using the given factor and long division. Use the Remainder Theorem to find the distance traveled after 45 seconds.

Polynomials

Well, we can also divide polynomials.

f(x) ÷ d(x) = q(x) with a remainder of r(x)

But it is better to write it as a sum like this:

Like in this example using Polynomial Long Division:

Example: 2x2−5x−1 divided by x−3

  • f(x) is 2x2−5x−1
  • d(x) is x−3

After dividing we get the answer 2x+1, but there is a remainder of 2.

  • q(x) is 2x+1
  • r(x) is 2

In the style f(x) = d(x)·q(x) + r(x) we can write:

2x2−5x−1 = (x−3)(2x+1) + 2

But you need to know one more thing:

Say we divide by a polynomial of degree 1 (such as 'x−3') the remainder will have degree 0 (in other words a constant, like '4').

We will use that idea in the 'Remainder Theorem':

The Remainder Theorem

Nonton film semi lies 1998. When we divide f(x) by the simple polynomial x−c we get:

f(x) = (x−c)·q(x) + r(x)

x−c is degree 1, so r(x) must have degree 0, so it is just some constant r:

f(x) = (x−c)·q(x) + r

Now see what happens when we have x equal to c:

f(c) =(0)·q(c) + r

So we get this:

The Remainder Theorem:

When we divide a polynomial f(x) by x−c the remainder is f(c)

So to find the remainder after dividing by x-c we don't need to do any division:

Just calculate f(c).

Let us see that in practice:

Example: The remainder after 2x2−5x−1 is divided by x−3

(Our example from above)

We don't need to divide by (x−3) .. just calculate f(3):

2(3)2−5(3)−1 = 2x9−5x3−1
= 18−15−1
= 2

And that is the remainder we got from our calculations above.

Meowanime sekirei s3 download. We didn't need to do Long Division at all!

Example: The remainder after 2x2−5x−1 is divided by x−5

Same example as above but this time we divide by 'x−5'

'c' is 5, so let us check f(5):

2(5)2−5(5)−1 = 2x25−5x5−1
= 50−25−1
= 24

The remainder is 24

Once again .. We didn't need to do Long Division to find that.

The Factor Theorem

Now ..

What if we calculate f(c) and it is 0?

.. that means the remainder is 0, and ..

.. (x−c) must be a factor of the polynomial!

We see this when dividing whole numbers. For example 60 ÷ 20 = 3 with no remainder. So 20 must be a factor of 60.

Example: x2−3x−4

f(4) = (4)2−3(4)−4 = 16−12−4 = 0

so (x−4) must be a factor of x2−3x−4

And so we have:

The Factor Theorem:

When f(c)=0 then x−c is a factor of f(x)

And the other way around, too:

When x−c is a factor of f(x) then f(c)=0

Why Is This Useful?

Knowing that x−c is a factor is the same as knowing that c is a root (and vice versa).

The factor 'x−c' and the root 'c' are the same thing

Know one and we know the other

For one thing, it means that we can quickly check if (x−c) is a factor of the polynomial.

Example: Find the factors of 2x3−x2−7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2. We can check easily:

f(2) = 2(2)3−(2)2−7(2)+2
= 16−4−14+2
= 0

Yes! f(2)=0, so we have found a root and a factor.

How about where it crosses near −1.8?

f(−1.8) = 2(−1.8)3−(−1.8)2−7(−1.8)+2
= −11.664−3.24+12.6+2
= −0.304

No, (x+1.8) is not a factor. We could try some other values near by and maybe get lucky.

But at least we know (x−2) is a factor, so let's use Polynomial Long Division:

2x2+3x−1
x−2)2x3− x2−7x+2
2x3−4x2
3x2−7x
3x2−6x
−x+2
−x+2
0

As expected the remainder is zero.

Better still, we are left with the quadratic equation2x2+3x−1 which is easy to solve.

It's roots are −1.78.. and 0.28.., so the final result is:

2x3−x2−7x+2 = (x−2)(x+1.78..)(x−0.28..)

Polynomial Factor Theorem Worksheet

We were able to solve a difficult polynomial.

Summary

The Remainder Theorem:

  • When we divide a polynomial f(x) by x−c the remainder is f(c)

The Factor Theorem:

  • When f(c)=0 then x−c is a factor of f(x)
  • When x−c is a factor of f(x) then f(c)=0

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.[1]

The factor theorem states that a polynomial f(x){displaystyle f(x)} has a factor (xk){displaystyle (x-k)}if and only iff(k)=0{displaystyle f(k)=0} (i.e. k{displaystyle k} is a root).[2]

Factorization of polynomials[edit]

Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.

The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:[3]

  1. 'Guess' a zero a{displaystyle a} of the polynomial f{displaystyle f}. (In general, this can be very hard, but maths textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
  2. Use the factor theorem to conclude that (xa){displaystyle (x-a)} is a factor of f(x){displaystyle f(x)}.
  3. Compute the polynomial g(x)=f(x)/(xa){displaystyle g(x)=f(x){big /}(x-a)}, for example using polynomial long division or synthetic division.
  4. Conclude that any root xa{displaystyle xneq a} of f(x)=0{displaystyle f(x)=0} is a root of g(x)=0{displaystyle g(x)=0}. Since the polynomial degree of g{displaystyle g} is one less than that of f{displaystyle f}, it is 'simpler' to find the remaining zeros by studying g{displaystyle g}.

Polynomial Remainder Theorem Calculator

Polynomial Factor Theorem Pdf

Example[edit]

Find the factors of

x3+7x2+8x+2.{displaystyle x^{3}+7x^{2}+8x+2.}

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if (x1){displaystyle (x-1)} is a factor, substitute x=1{displaystyle x=1} into the polynomial above:

x3+7x2+8x+2=(1)3+7(1)2+8(1)+2{displaystyle x^{3}+7x^{2}+8x+2=(1)^{3}+7(1)^{2}+8(1)+2}
=1+7+8+2{displaystyle =1+7+8+2}
=18.{displaystyle =18.}

As this is equal to 18 and not 0 this means (x1){displaystyle (x-1)} is not a factor of x3+7x2+8x+2{displaystyle x^{3}+7x^{2}+8x+2}. So, we next try (x+1){displaystyle (x+1)} (substituting x=1{displaystyle x=-1} into the polynomial):

(1)3+7(1)2+8(1)+2.{displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2.}

This is equal to 0{displaystyle 0}. Therefore x(1){displaystyle x-(-1)}, which is to say x+1{displaystyle x+1}, is a factor, and 1{displaystyle -1} is a root of x3+7x2+8x+2.{displaystyle x^{3}+7x^{2}+8x+2.}

The next two roots can be found by algebraically dividing x3+7x2+8x+2{displaystyle x^{3}+7x^{2}+8x+2} by (x+1){displaystyle (x+1)} to get a quadratic:

x3+7x2+8x+2x+1=x2+6x+2,{displaystyle {x^{3}+7x^{2}+8x+2 over x+1}=x^{2}+6x+2,}

and therefore (x+1){displaystyle (x+1)} and x2+6x+2{displaystyle x^{2}+6x+2} are factors of x3+7x2+8x+2.{displaystyle x^{3}+7x^{2}+8x+2.} Of these the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic 3±7.{displaystyle -3pm {sqrt {7}}.} Thus the three irreducible factors of the original polynomial are x+1,{displaystyle x+1,}x(3+7),{displaystyle x-(-3+{sqrt {7}}),} and x(37).{displaystyle x-(-3-{sqrt {7}}).}

Polynomial Factor Theorem Pdf Download

References[edit]

Polynomial Factor Theorem Pdf

  1. ^Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN0-13-370149-2.
  2. ^Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN978-81-317-2816-1.
  3. ^Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN81-7008-629-9.

Division Theorem For Polynomials

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