Quick Polynomial Question (1 Viewer)

[nrlwinner]

You are given that the polynomial has a double root. Find all the roots using the idea of multiplicity of roots.

 

[life92]

P(x)=4x^3+12x^2-15x+4

P'(x)=12x^2+24x-15
= 3 (4x^2+8x-5)
= 3 (2x+5)(2x-1)

P(1/2) = 0 , therefore the double root is 1/2, or the double factor is (2x-1)
By inspection, the last factor must then be (x+4)

Therefore the roots are 1/2, 1/2 and -4

 

[nrlwinner]

I know this may seem like a simple question, but...

For



You are told that it has a rational root. The answer says its 3/2 but how can you do this using the factor theorem?

 

[cutemouse]

I know this may seem like a simple question, but...

For



You are told that it has a rational root. The answer says its 3/2 but how can you do this using the factor theorem?

Let the rational root be of form r/s where r|3 and s|2 and r and s are relatively prime.

Thus possible rational roots are: 1/1, 3/1, 1/2 and 3/2. etc

 

[nrlwinner]

Let the rational root be of form r/s where r|3 and s|2 and r and s are relatively prime.

Thus possible rational roots are: 1/1, 3/1, 1/2 and 3/2. etc

Could you please explain this.

 

[cutemouse]

It's a basic rule.

If a polynomial, say, P(x)=ax^3+bx^2+cx+d has a rational root, then it is of the form x=r/s where r|d and s|a and where r and s are relatively prime.

 

[untouchablecuz]

It's a basic rule.

If a polynomial, say, P(x)=ax^3+bx^2+cx+d has a rational root, then it is of the form x=r/s where r|d and s|a and where r and s are relatively prime.

he doesn't understand the notation etc

basically:

P(x)=a_nxⁿ+a_n-1x^n-1+...+a₁x+a₀

if P(x) has real and rational root z, then z can be expressed in the form p/q, where p is a divisor of a₀ and q is a divisor of the leading co-efficient a_n

 

[nrlwinner]

Got another question from Cambridge that I'm stuck on. PS. I looked at the solutions, but I still don't get it. I need someone to explain it to me step by step.

P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root2, another zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

 

[cutemouse]

P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root2, another zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

Let P(x)=ax^4+bx^3+cx^2+dx+e

Constant term is 4, thus e=4
P(x) is monic, thus a=1

P(x)=x^4+bx^3+cx^2+dx+4

If one zero is \sqrt2

then



As P(x) has rational (integer) coeff, then -\sqrt2 is also a zero.

Thus we have



Hmm I need one more bit of information... Will finish later.

 

[untouchablecuz]

Got another question from Cambridge that I'm stuck on. PS. I looked at the solutions, but I still don't get it. I need someone to explain it to me step by step.

P(x) is a monic polynomial of degree 4 with integer coefficients and constant term 4. One zero is root2, another zero is rational and the sum of the zeros is positive. Factorise P(x) fully over R.

i answered that here:

http://community.boredofstudies.org/14/mathematics-extension-2/239096/polynomials.html#post4943090