Answer :
Certainly! Let's solve each part of the question step-by-step.
### Part (a)
To find the remainder when [tex]\( f(x) = 6x^3 + 13x^2 - 4 \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] using the remainder theorem, we do the following:
1. According to the remainder theorem, when a polynomial [tex]\( f(x) \)[/tex] is divided by a linear binomial of the form [tex]\( ax + b \)[/tex], the remainder of that division is [tex]\( f\left(-\frac{b}{a}\right) \)[/tex].
2. For the binomial [tex]\( 2x + 3 \)[/tex], we set [tex]\( 2x + 3 = 0 \)[/tex] to find the value of [tex]\( x \)[/tex]:
[tex]\[ 2x + 3 = 0 \implies x = -\frac{3}{2} \][/tex]
3. Substitute [tex]\( x = -\frac{3}{2} \)[/tex] into the polynomial [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(-\frac{3}{2}\right) = 6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4 \][/tex]
4. Calculate the value:
[tex]\[ \left(-\frac{3}{2}\right)^3 = -\frac{27}{8} \][/tex]
[tex]\[ 6 \cdot -\frac{27}{8} = -\frac{162}{8} = -\frac{81}{4} \][/tex]
[tex]\[ \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \][/tex]
[tex]\[ 13 \cdot \frac{9}{4} = \frac{117}{4} \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = -\frac{81}{4} + \frac{117}{4} - 4 \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = \frac{36}{4} - 4 = 9 - 4 = 5 \][/tex]
So, the remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] is [tex]\( \boxed{5} \)[/tex].
### Part (b)
To show that [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex] using the factor theorem, we do the following:
1. According to the factor theorem, [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex] if and only if [tex]\( f(-2) = 0 \)[/tex].
2. Substitute [tex]\( x = -2 \)[/tex] into the polynomial [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = 6(-2)^3 + 13(-2)^2 - 4 \][/tex]
3. Calculate the value:
[tex]\[ (-2)^3 = -8 \][/tex]
[tex]\[ 6 \cdot -8 = -48 \][/tex]
[tex]\[ (-2)^2 = 4 \][/tex]
[tex]\[ 13 \cdot 4 = 52 \][/tex]
[tex]\[ f(-2) = -48 + 52 - 4 = 0 \][/tex]
Since [tex]\( f(-2) = 0 \)[/tex], this confirms that [tex]\( x + 2 \)[/tex] is indeed a factor of [tex]\( f(x) \)[/tex].
### Part (c)
To factorize [tex]\( f(x) \)[/tex] completely, we start by using the factor we already know, [tex]\( x + 2 \)[/tex], and then find other factors.
1. We already know that [tex]\( f(x) = 6x^3 + 13x^2 - 4 \)[/tex] has a factor [tex]\( x + 2 \)[/tex].
2. Next, we divide [tex]\( f(x) \)[/tex] by [tex]\( x + 2 \)[/tex].
By dividing [tex]\( 6x^3 + 13x^2 - 4 \)[/tex] by [tex]\( x + 2 \)[/tex], we find the other factors to get the completely factored form:
[tex]\[ f(x) = (x + 2)(2x - 1)(3x + 2) \][/tex]
Thus, the completely factorized form of [tex]\( f(x) \)[/tex] is [tex]\( \boxed{(x+2)(2x-1)(3x+2)} \)[/tex].
To summarize:
- Part (a): Remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] is [tex]\( \boxed{5} \)[/tex].
- Part (b): [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex].
- Part (c): The completely factorized form of [tex]\( f(x) \)[/tex] is [tex]\( \boxed{(x+2)(2x-1)(3x+2)} \)[/tex].
### Part (a)
To find the remainder when [tex]\( f(x) = 6x^3 + 13x^2 - 4 \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] using the remainder theorem, we do the following:
1. According to the remainder theorem, when a polynomial [tex]\( f(x) \)[/tex] is divided by a linear binomial of the form [tex]\( ax + b \)[/tex], the remainder of that division is [tex]\( f\left(-\frac{b}{a}\right) \)[/tex].
2. For the binomial [tex]\( 2x + 3 \)[/tex], we set [tex]\( 2x + 3 = 0 \)[/tex] to find the value of [tex]\( x \)[/tex]:
[tex]\[ 2x + 3 = 0 \implies x = -\frac{3}{2} \][/tex]
3. Substitute [tex]\( x = -\frac{3}{2} \)[/tex] into the polynomial [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(-\frac{3}{2}\right) = 6\left(-\frac{3}{2}\right)^3 + 13\left(-\frac{3}{2}\right)^2 - 4 \][/tex]
4. Calculate the value:
[tex]\[ \left(-\frac{3}{2}\right)^3 = -\frac{27}{8} \][/tex]
[tex]\[ 6 \cdot -\frac{27}{8} = -\frac{162}{8} = -\frac{81}{4} \][/tex]
[tex]\[ \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \][/tex]
[tex]\[ 13 \cdot \frac{9}{4} = \frac{117}{4} \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = -\frac{81}{4} + \frac{117}{4} - 4 \][/tex]
[tex]\[ f\left(-\frac{3}{2}\right) = \frac{36}{4} - 4 = 9 - 4 = 5 \][/tex]
So, the remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] is [tex]\( \boxed{5} \)[/tex].
### Part (b)
To show that [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex] using the factor theorem, we do the following:
1. According to the factor theorem, [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex] if and only if [tex]\( f(-2) = 0 \)[/tex].
2. Substitute [tex]\( x = -2 \)[/tex] into the polynomial [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = 6(-2)^3 + 13(-2)^2 - 4 \][/tex]
3. Calculate the value:
[tex]\[ (-2)^3 = -8 \][/tex]
[tex]\[ 6 \cdot -8 = -48 \][/tex]
[tex]\[ (-2)^2 = 4 \][/tex]
[tex]\[ 13 \cdot 4 = 52 \][/tex]
[tex]\[ f(-2) = -48 + 52 - 4 = 0 \][/tex]
Since [tex]\( f(-2) = 0 \)[/tex], this confirms that [tex]\( x + 2 \)[/tex] is indeed a factor of [tex]\( f(x) \)[/tex].
### Part (c)
To factorize [tex]\( f(x) \)[/tex] completely, we start by using the factor we already know, [tex]\( x + 2 \)[/tex], and then find other factors.
1. We already know that [tex]\( f(x) = 6x^3 + 13x^2 - 4 \)[/tex] has a factor [tex]\( x + 2 \)[/tex].
2. Next, we divide [tex]\( f(x) \)[/tex] by [tex]\( x + 2 \)[/tex].
By dividing [tex]\( 6x^3 + 13x^2 - 4 \)[/tex] by [tex]\( x + 2 \)[/tex], we find the other factors to get the completely factored form:
[tex]\[ f(x) = (x + 2)(2x - 1)(3x + 2) \][/tex]
Thus, the completely factorized form of [tex]\( f(x) \)[/tex] is [tex]\( \boxed{(x+2)(2x-1)(3x+2)} \)[/tex].
To summarize:
- Part (a): Remainder when [tex]\( f(x) \)[/tex] is divided by [tex]\( 2x + 3 \)[/tex] is [tex]\( \boxed{5} \)[/tex].
- Part (b): [tex]\( x + 2 \)[/tex] is a factor of [tex]\( f(x) \)[/tex].
- Part (c): The completely factorized form of [tex]\( f(x) \)[/tex] is [tex]\( \boxed{(x+2)(2x-1)(3x+2)} \)[/tex].
