To factorize the polynomial [tex]\( 2x^3 - 7x^2 - 10x + 24 \)[/tex] using synthetic division, we will follow these steps:
1. Find a root of the polynomial:
We can use the Rational Root Theorem, which states that any potential rational root of the polynomial [tex]\( 2x^3 - 7x^2 - 10x + 24 \)[/tex] is a factor of the constant term (24) divided by a factor of the leading coefficient (2). Possible rational roots are [tex]\( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \)[/tex] divided by [tex]\( \pm 1, \pm 2 \)[/tex].
2. Test the potential roots:
Trying [tex]\( x = 4 \)[/tex]:
[tex]\[
2(4)^3 - 7(4)^2 - 10(4) + 24 = 128 - 112 - 40 + 24 = 0
\][/tex]
So, [tex]\( x = 4 \)[/tex] is a root.
3. Use synthetic division to factor the polynomial:
Now that we have a root, [tex]\( x - 4 \)[/tex] is a factor. We use synthetic division to divide the polynomial by [tex]\( x - 4 \)[/tex].
The coefficients of the polynomial [tex]\( 2x^3 - 7x^2 - 10x + 24 \)[/tex] are [tex]\( 2, -7, -10, 24 \)[/tex].
Perform synthetic division with [tex]\( x = 4 \)[/tex]:
[tex]\[
\begin{array}{r|rrrr}
4 & 2 & -7 & -10 & 24 \\
\hline
& 2 & 1 & -6 & 0 \\
\end{array}
\][/tex]
The process yields [tex]\( 2, 1, -6 \)[/tex], giving us [tex]\( 2x^2 + x - 6 \)[/tex] as the quotient.
4. Factor the quotient:
Next, factor the quadratic polynomial [tex]\( 2x^2 + x - 6 \)[/tex]. We look for two numbers that multiply to [tex]\( 2 \times (-6) = -12 \)[/tex] and add to [tex]\( 1 \)[/tex].
These numbers are [tex]\( 3 \)[/tex] and [tex]\( -4 \)[/tex]. So,
[tex]\[
2x^2 + x - 6 = 2x^2 + 3x - 2x - 6 = x(2x + 3) - 2(2x + 3) = (x - 2)(2x + 3)
\][/tex]
5. Write the final factorization:
Combining all factors, we get:
[tex]\[
2x^3 - 7x^2 - 10x + 24 = (x - 4)(x + 2)(2x - 3)
\][/tex]
Thus, the factorization of the polynomial [tex]\( 2x^3 - 7x^2 - 10x + 24 \)[/tex] is:
[tex]\[
2x^3 - 7x^2 - 10x + 24 = (x-4)(x+2)(2x-3)
\][/tex]
