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:

$p(x) = (3x^{2} - 8x + 5)(x - 4)$

Step-by-step explanation:

p(x) has degree 3

(x - 4) has degree 1

So p(x) can be rewritten as a polynomial of degree 2 multiplying (x-4)

So

$p(x) = (ax^{2} + bx + c)(x - 4)$

To find a, b and c, we have to divide p(x) by x - 4.

So

Finding a:

Dividing the first term of p(x) by x - 4.

$\frac{3x^{3}}{x - 4} = 3x^{2}$

So a = 3.

Now multiplying 3x² by, x - 4, we have:

$3x^{2}(x - 4) = 3x^{3} - 12x^{2}$

Subtracting p(x) from this:

$3x^{3} - 20x^{2} + 37x - 20 - (3x^{3} - 12x^{2}) = 3x^{3} - 20x^{2} + 37x - 20 - 3x^{3} + 12x^{2} = -8x^{2} + 37x - 20$

Finding b:

Dividing the first term, after the subtraction, by x - 4.

$\frac{-8x^{2}}{x - 4} = -8x$

So b = -8.

Multiplying -8x by x - 4, we have:

$-8x(x-4) = -8x^{2} + 32x$

Then

$-8x^{2} + 37x - 20 - (-8x^{2} + 32x) = -8x^{2} + 37x - 20 + 8x^{2} - 32x = 5x - 20$

Finding c:

$\frac{5x}{x - 4} = 5$

So c = 5.

Just to verify if the remainder is 0.

$5(x - 4) = 5x - 20$

$5x - 20 - (5x - 20) = 0$

Ok

Then:

$p(x) = (ax^{2} + bx + c)(x - 4)$

$p(x) = (3x^{2} - 8x + 5)(x - 4)$