Question

Find a binomial that is a factor of this polynomial (4x^2 - 2x + 14)

Answer 1

There aren't any binomial factors: no real roots exist.

Answer 2

do you mean in this equation or in general because then that would be impossible

Answer 3

There are binomial factors they're just not real... the given quadratic is irreducible over the reals. Generally, @PinkPanda, if some real \(x_0\) exists s.t. \(P(x_0)=0\) for some polynomial function \(P(x)\) then \(x-x_0\) is a binomial factor.

Answer 4

so how am i suppose to divide a binomial that is a factor of that polynomial? @oldrin.bataku

Answer 5

What do you mean by that? and the given polynomial cannot be factored because it has no real roots/zeroes (graphically, the parabola it describes does not touch the \(x\)-axis).

Answer 6

ok this is what im supposed to do for math :Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial? Create a sample problem that has a binomial which IS a factor of the polynomial being divided, and another problem that has a binomial which is NOT a factor of the polynomial being divided. @oldrin.bataku

Answer 7

You can determine whether a binomial \(x-c\) is a factor of a polynomial \(P\) in \(x\) by verifying that \(P(c)=0\).

Answer 8

im so lost

Answer 9

This should make intuitive sense; if \(x-c_1,x-c_2,x-c_3,\dots,x-c_n\) are the \(n\) factors of \(P\) then we can write \(P(x)=(x-c_1)(x-c_2)(x-c_3)\cdots(x-c_n)\). Plugging in any of \(c_1,c_2,c_3,\cdots,c_n\) for \(x\) will make one of the factors become \(0\) so that the whole thing becomes \(0\) -- these are what roots/zeroes are.

Answer 10

ok??? so how do i make a polynomial that can have a binomial that is a factor
im sorry but this confusing