Question $P(x)=2 x^{3}-15 x^{2}+27 x-10 \quad x_{1}=5$

Studdy Solution

STEP 1

1. The polynomial given is $P(x) = 2x^3 - 15x^2 + 27x - 10$ .
2. We are given a root $x_1 = 5$ , meaning $P(5) = 0$ .
3. We aim to use polynomial division to factor $P(x)$ by $(x-5)$ .
4. After factoring out $(x-5)$ , we will obtain a quadratic factor which can be further factored or solved using the quadratic formula.

STEP 2

1. Verify that $x=5$ is a root of $P(x)$ .
2. Perform polynomial division of $P(x)$ by $(x-5)$ .
3. Simplify the polynomial division to find the quadratic quotient.
4. Factor the quadratic polynomial or solve for its roots.

STEP 3

Verify that $x=5$ is a root of $P(x)$ by substituting $x=5$ into $P(x)$ .
$P(5) = 2(5)^3 - 15(5)^2 + 27(5) - 10$ $P(5) = 2 \cdot 125 - 15 \cdot 25 + 135 - 10$ $P(5) = 250 - 375 + 135 - 10$ $P(5) = 0$
Since $P(5) = 0$ , $x=5$ is indeed a root.

STEP 4

Perform polynomial division of $P(x)$ by $(x-5)$ to find the quotient polynomial.
$\text{Divide } 2x^3 - 15x^2 + 27x - 10 \text{ by } x-5$

STEP 5

Set up the polynomial division.
$\begin{array}{r|rrrr} x-5 & 2x^3 & -15x^2 & +27x & -10 \\ \hline & 2x^2 & -5x & +2 & \\ \end{array}$
1. Divide the leading term $2x^3$ by $x$ to get $2x^2$ .
2. Multiply $2x^2$ by $(x-5)$ and subtract the result from $2x^3 - 15x^2 + 27x - 10$ .
3. Repeat for the remaining terms.

STEP 6

Continuing the polynomial division step-by-step:
1. Multiply $2x^2$ by $(x-5)$ :
$2x^2(x-5) = 2x^3 - 10x^2$
2. Subtract from the original polynomial:
$(2x^3 - 15x^2 + 27x - 10) - (2x^3 - 10x^2) = -5x^2 + 27x - 10$
3. Repeat the process with $-5x$ :
$-5x(x-5) = -5x^2 + 25x$
4. Subtract again:
$(-5x^2 + 27x - 10) - (-5x^2 + 25x) = 2x - 10$
5. Finally, divide $2x - 10$ by $x-5$ :
$2(x-5) = 2x - 10$
The quotient is $2x^2 - 5x + 2$ .

STEP 7

The quotient polynomial is $2x^2 - 5x + 2$ . We now factor this quadratic polynomial.
Solve $2x^2 - 5x + 2 = 0$ using the quadratic formula where $a=2$ , $b=-5$ , and $c=2$ .
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 8

Calculate the discriminant:
$b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9$
So, the roots are:
$x = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}$
This gives us two roots:
$x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{2}{4} = \frac{1}{2}$

STEP 9

The complete factorization of $P(x)$ is:
$P(x) = 2(x-5)(x-2)\left(x-\frac{1}{2}\right)$
Thus, the roots of the polynomial are $x = 5$ , $x = 2$ , and $x = \frac{1}{2}$ .

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