Find all the zeros of h(x)= 7x^3+13x^2+19x-3 given that it has at least one rational zero?

h(x) = 7x^3 + 13x^2 + 19x - 3

We know that it has one rational zero, so all we have to do is find all factors of the coefficient of x^3, and all factors of the constant term.

Factors of -3: 1, -1, 3, -3

Factors of 7: 1, -1, 7, -7

Put all factors of -3 divided by all factors of 7. These will be our possible roots.

(1, -1, 3, -3) / (1, -1, 7, -7) =

1/1, 1/(-1), 7/1, -7/1, -1/1, -1/(-1), -1/7, -1/(-7), 3/1, 3/(-1), 3/7,

3/(-7), -3/1, -3/(-1), -3/7, -3/(-7)

But some of them are repeat terms; the possible rational roots are:

1, -1, 7, -7, -1/7, 1/7, 3, -3, 3/7, -3/7

Test each one for zero. Once you find one that works (which we'll call r), it follows that (x - r) will be a factor, and you can use synthetic long division to get the other zeros.

According to the rational roots theorem, if a rational r = m/n, m and n<> integers to their lowest terms, is a rot of h, then m divedes -3 (the independent coefficient) and n divides 7 (the coefficient of the leading term). So, the only possibilites for m are 1, -1, 3, -3 and the only possibilities for n are 1 , -1, 7 , -7. Combining such values, the possibilities for r are 1, -1, 1/7, -1/7, 3 , -3 , 3/7 , -3/7

We readily see that 7, 1 , 3 and -1 aren't roots. So what you have to do is test the other possible roots. If the statement is true, for at least one you'll have p(r) =0. As soon as you find one, you can divide h by x -r, getting a 2nd polynomial whose roots are the other roots of h. Then, apply Bhaskara formula nd get the other roots.

To do this one, I'll consider all possible rational zeroes, from the rational roots theorem, then use synthetic division to find one. This will break it down to a quadratic, whose zeros can be found via factoring or the quad formula

Ok, by the rational roots theorem, the rational roots have to be

+/- 3, +/- 3/7, or +/- 1/7

http://mathforum.org/library/drmath/view/56425.htm...

By synthetic division we can find that 1/7 is a root, and so (7x - 1) is a factor.

http://www.purplemath.com/modules/synthdiv.htm

Or, if you'd rather not do synthetic division with a fraction divisor, just divide (7x - 1) into the given expression.

http://www.purplemath.com/modules/polydiv2.htm

Either way, you'll find that the quotient is x^2 + x + 3

This does not factor, so put it in the quad formula and you'll find that the other two roots are complex numbers,

(-1 +/- i sqrt 11)/2

If you want only real solutions you'd discard these complex solutions and just say the zero is 1/7

http://mathworld.wolfram.com/RationalZeroTheorem.h...

The rational root must have a factor of 3 as its numerator and a factor of 7 as its denominator. After trying 3/7, -3/7, and other possibilities, you find that 1/7 is a root.

Divide the polynomial by (x - 1/7), then solve the quadratic to find the other two roots.

h(x)=7x3+13x2+19x-3

7x3+13x2+19x-3 =7x3-x2+14x2-2x+21x-3 =(7x3-x2)+(14x2-2x)+(21x-3) =

= (7x-1)x2+(7x-12)x+(7x-1)3 =

=(7x - 1)(x2 + x +3)

7x-1= 0 => x=1/7

x2+x+3=0

D<0 => no real roots

x=1/7 - zero