Question

What is factor theorem

Answer 1

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 {\displaystyle f(x)} has a factor {\displaystyle (x-k)}if and only if {\displaystyle f(k)=0} (i.e. {\displaystyle k} is a root)

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]

"Guess" a zero {\displaystyle a} of the polynomial {\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.)Use the factor theorem to conclude that {\displaystyle (x-a)} is a factor of {\displaystyle f(x)}.Compute the polynomial {\displaystyle g(x)=f(x){\big /}(x-a)}, for example usingpolynomial long division or synthetic division.Conclude that any root {\displaystyle x\neq a} of {\displaystyle f(x)=0}is a root of {\displaystyle g(x)=0}. Since the polynomial degree of {\displaystyle g} is one less than that of {\displaystyle f}, it is "simpler" to find the remaining zeros by studying {\displaystyle g}.

ExampleEdit

Find the factors at

{\displaystyle x^{3}+7x^{2}+8x+2.}

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

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

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

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

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

The next two roots can be found by algebraically dividing {\displaystyle x^{3}+7x^{2}+8x+2} by{\displaystyle (x+1)} to get a quadratic, which can be solved directly, by the factor theorem or by thequadratic formula.

{\displaystyle {x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2}

and therefore {\displaystyle (x+1)} and {\displaystyle x^{2}+6x+2} are the factors of {\displaystyle x^{3}+7x^{2}+8x+2.}

Answer 2

If p (x) is a polynomial og degree n > 1 and a is any real number ,then
X-a is a factor of p(x) , if p(a) =0
p (a)= 0 ,if x-a is a factor of p (x)

hope this will help u