Question

The polynomial p(x)=3x^3-20x^2+37x-20 has a known factor of (x-4). Rewrite p(x) as a product of linear factors. p(x) =

Answer 1

Answer:

$p(x) = (x-4)(x - \frac{5}{3})(x - 1)$

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

We have the polynomial $p(x) = 3x^3-20x^2+37x-20$

It has a known factor $x - 4$. This means that the polynomial can be written as:

$p(x) = (x-4)q(x)$

In which q(x) is a second order polynomial, because p is of the third degree and q is of the first degree(3 - 1 = 2). So

$q(x) = ax^2 + bx + c$

We have to find a, b and c. Then

$3x^3-20x^2+37x-20 = (x-4)(ax^2 + bx + c)$

$3x^3-20x^2+37x-20 = = ax^3 + (-4a + b)x^2 + (c - 4b)x - 4c$

Comparing both sides, we have that

$a = 3$

$-4c = -20 \rightarrow c = 5$

$c - 4b = 37$

$-4b = 32 \rightarrow b = -8$

Now we find the roots of this polynomial.

$\bigtriangleup = b^{2} - 4ac = (-8)^2 - 4(3)(5) = 64 - 60 = 4$

$x_{1} = \frac{-(-8) + \sqrt{4}}{2*3} = \frac{5}{3}$

$x_{2} = \frac{-(-8) - \sqrt{4}}{2*3} = 1$

So, as a product of it's factors, we have that q is:

$q(x) = (x - \frac{5}{3})(x - 1)$

And p(x)

$p(x) = (x-4)q(x)$

$p(x) = (x-4)(x - \frac{5}{3})(x - 1)$

Answer 2

Answer: (x-4)(3x-5)(x-1)

Step-by-step explanation: