Thread: Factoring polynomials.

1) x=(x^3)-(3x^2)-13x+15

Its zeros are ?,?,? (3 answers)


2)The polynomial of degree 4, p(x) has a root of multiplicity 2 at x=2 and roots of multiplicity 1 at x=0 and x=-2. Its lead coefficient is 5.
Find a formula for . p(x)

Originally Posted by wwinston

1) x=(x^3)-(3x^2)-13x+15

Its zeros are ?,?,? (3 answers)

$\displaystyle \displaystyle x=x^3-3x^2-13x+15$

$\displaystyle \displaystyle 0=x^3-3x^2-14x+15$

Use the factor theorem with factors of 15 to check for integer solutions.

Originally Posted by wwinston

1) x=(x^3)-(3x^2)-13x+15

Its zeros are ?,?,? (3 answers)


2)The polynomial of degree 4, p(x) has a root of multiplicity 2 at x=2 and roots of multiplicity 1 at x=0 and x=-2. Its lead coefficient is 5.
Find a formula for . p(x)

did you mean $\displaystyle f(x) = x^3 - 3x^2 - 13x + 15$ ? if so, note that $\displaystyle f(1) = 0$ ... you should now be able to find the other two zeros.


look at the following polynomial in factored form ...

$\displaystyle p(x) = ax(x-b)(x-c)^2$

it has a leading coefficient $\displaystyle a$ , root at $\displaystyle x = 0$ , root at $\displaystyle x = b$ , and a root of multiplicity two at $\displaystyle x = c$

Originally Posted by wwinston

1) x=(x^3)-(3x^2)-13x+15

You must mean this: x^3 - 3x^2 - 13x + 15 = 0 ; right?
You can tell easily that x=1 is a solution; so do a division by x-1 next.

yeah thats what I meant...

i factored and got

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

figuring the answers would be 0,2,1

the only one that is right is 1..i can check the answers to see if im right unlimited times

$\displaystyle (x^3 - 3x^2 - 13x + 15)\div (x-1) = x^2-2x-15 $