Answer :
Let's solve the given problem step-by-step:
Given the polynomial function:
[tex]\[ f(x) = 2x^3 - 5x^2 + ax + 18 \][/tex]
where [tex]\( a \)[/tex] is a constant, and it is given that [tex]\((x - 3)\)[/tex] is a factor of [tex]\( f(x) \)[/tex].
### Part (a) Show that [tex]\( a = -9 \)[/tex]
1. Factor Theorem:
- The Factor Theorem states that if [tex]\( (x - c) \)[/tex] is a factor of a polynomial [tex]\( f(x) \)[/tex], then [tex]\( f(c) = 0 \)[/tex].
- Since [tex]\( (x - 3) \)[/tex] is a factor, [tex]\( f(3) = 0 \)[/tex].
2. Substitute [tex]\( x = 3 \)[/tex] in [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{align*} f(x) &= 2x^3 - 5x^2 + ax + 18, \\ f(3) &= 2(3)^3 - 5(3)^2 + a(3) + 18. \end{align*} \][/tex]
3. Calculate [tex]\( f(3) \)[/tex]:
[tex]\[ \begin{align*} 2(3)^3 &= 2 \cdot 27 = 54, \\ -5(3)^2 &= -5 \cdot 9 = -45, \\ a(3) &= 3a, \\ 54 - 45 + 3a + 18 &= 0. \end{align*} \][/tex]
4. Simplify the equation:
[tex]\[ 54 - 45 + 18 + 3a = 0, \\ 27 + 3a = 0. \][/tex]
5. Solve for [tex]\( a \)[/tex]:
[tex]\[ 3a = -27, \\ a = -9. \][/tex]
Hence, [tex]\( a = -9 \)[/tex].
### Part (b) Factorise [tex]\( f(x) \)[/tex] completely
1. Substitute [tex]\( a = -9 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2x^3 - 5x^2 - 9x + 18. \][/tex]
2. Verify that [tex]\( (x - 3) \)[/tex] is a factor:
- We already know that [tex]\( f(3) = 0 \)[/tex], therefore, [tex]\( (x - 3) \)[/tex] is indeed a factor.
3. Perform polynomial division or factor by grouping:
We factor the polynomial by recognizing that [tex]\( (x - 3) \)[/tex] is a factor. We use the quotient obtained from dividing [tex]\( 2x^3 - 5x^2 - 9x + 18 \)[/tex] by [tex]\( (x - 3) \)[/tex].
4. Divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 3) \)[/tex]:
[tex]\[ f(x) \div (x-3) = 2x^3 - 5x^2 - 9x + 18 \div (x-3) \][/tex]
Using synthetic or polynomial division (details omitted for brevity), we get:
[tex]\[ = (x - 3)(2x^2 + x - 6). \][/tex]
5. Factorise the quadratic factor [tex]\( 2x^2 + x - 6 \)[/tex]:
To factor [tex]\( 2x^2 + x - 6 \)[/tex], we look for two numbers that multiply to [tex]\( 2 \cdot (-6) = -12 \)[/tex] and add up to [tex]\( 1 \)[/tex]:
[tex]\[ 2x^2 + 4x - 3x - 6 \Rightarrow \text{Group terms} \Rightarrow 2x(x + 2) - 3(x + 2). \][/tex]
Factoring by grouping:
[tex]\[ = (2x - 3)(x + 2). \][/tex]
6. Combine all factors:
[tex]\[ f(x) = (x - 3)(2x^2 + x - 6) = (x - 3)(2x - 3)(x + 2). \][/tex]
Therefore, the completely factorized form of [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = (x - 3)(x + 2)(2x - 3). \][/tex]
Given the polynomial function:
[tex]\[ f(x) = 2x^3 - 5x^2 + ax + 18 \][/tex]
where [tex]\( a \)[/tex] is a constant, and it is given that [tex]\((x - 3)\)[/tex] is a factor of [tex]\( f(x) \)[/tex].
### Part (a) Show that [tex]\( a = -9 \)[/tex]
1. Factor Theorem:
- The Factor Theorem states that if [tex]\( (x - c) \)[/tex] is a factor of a polynomial [tex]\( f(x) \)[/tex], then [tex]\( f(c) = 0 \)[/tex].
- Since [tex]\( (x - 3) \)[/tex] is a factor, [tex]\( f(3) = 0 \)[/tex].
2. Substitute [tex]\( x = 3 \)[/tex] in [tex]\( f(x) \)[/tex]:
[tex]\[ \begin{align*} f(x) &= 2x^3 - 5x^2 + ax + 18, \\ f(3) &= 2(3)^3 - 5(3)^2 + a(3) + 18. \end{align*} \][/tex]
3. Calculate [tex]\( f(3) \)[/tex]:
[tex]\[ \begin{align*} 2(3)^3 &= 2 \cdot 27 = 54, \\ -5(3)^2 &= -5 \cdot 9 = -45, \\ a(3) &= 3a, \\ 54 - 45 + 3a + 18 &= 0. \end{align*} \][/tex]
4. Simplify the equation:
[tex]\[ 54 - 45 + 18 + 3a = 0, \\ 27 + 3a = 0. \][/tex]
5. Solve for [tex]\( a \)[/tex]:
[tex]\[ 3a = -27, \\ a = -9. \][/tex]
Hence, [tex]\( a = -9 \)[/tex].
### Part (b) Factorise [tex]\( f(x) \)[/tex] completely
1. Substitute [tex]\( a = -9 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2x^3 - 5x^2 - 9x + 18. \][/tex]
2. Verify that [tex]\( (x - 3) \)[/tex] is a factor:
- We already know that [tex]\( f(3) = 0 \)[/tex], therefore, [tex]\( (x - 3) \)[/tex] is indeed a factor.
3. Perform polynomial division or factor by grouping:
We factor the polynomial by recognizing that [tex]\( (x - 3) \)[/tex] is a factor. We use the quotient obtained from dividing [tex]\( 2x^3 - 5x^2 - 9x + 18 \)[/tex] by [tex]\( (x - 3) \)[/tex].
4. Divide [tex]\( f(x) \)[/tex] by [tex]\( (x - 3) \)[/tex]:
[tex]\[ f(x) \div (x-3) = 2x^3 - 5x^2 - 9x + 18 \div (x-3) \][/tex]
Using synthetic or polynomial division (details omitted for brevity), we get:
[tex]\[ = (x - 3)(2x^2 + x - 6). \][/tex]
5. Factorise the quadratic factor [tex]\( 2x^2 + x - 6 \)[/tex]:
To factor [tex]\( 2x^2 + x - 6 \)[/tex], we look for two numbers that multiply to [tex]\( 2 \cdot (-6) = -12 \)[/tex] and add up to [tex]\( 1 \)[/tex]:
[tex]\[ 2x^2 + 4x - 3x - 6 \Rightarrow \text{Group terms} \Rightarrow 2x(x + 2) - 3(x + 2). \][/tex]
Factoring by grouping:
[tex]\[ = (2x - 3)(x + 2). \][/tex]
6. Combine all factors:
[tex]\[ f(x) = (x - 3)(2x^2 + x - 6) = (x - 3)(2x - 3)(x + 2). \][/tex]
Therefore, the completely factorized form of [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = (x - 3)(x + 2)(2x - 3). \][/tex]
