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Sagot :
Sure, let's solve each part step-by-step.
### 3a). Find the remainder when [tex]\(x^3 + x^2 + x - 2\)[/tex] is divided by [tex]\(x + 3\)[/tex]
We will use the Remainder Theorem for this. The Remainder Theorem states that the remainder of the division of a polynomial [tex]\( f(x) \)[/tex] by a linear divisor [tex]\( x - a \)[/tex] is [tex]\( f(a) \)[/tex].
In this case, we are dividing [tex]\( f(x) = x^3 + x^2 + x - 2 \)[/tex] by [tex]\( x + 3 \)[/tex]. Notice that [tex]\( x + 3 \)[/tex] can be written as [tex]\( x - (-3) \)[/tex]. Hence, [tex]\( a = -3 \)[/tex].
1. Substitute [tex]\( x = -3 \)[/tex] into the polynomial:
[tex]\[ f(-3) = (-3)^3 + (-3)^2 + (-3) - 2 \][/tex]
2. Calculate each term:
[tex]\[ (-3)^3 = -27 \][/tex]
[tex]\[ (-3)^2 = 9 \][/tex]
[tex]\[ (-3) = -3 \][/tex]
[tex]\[ -2 = -2 \][/tex]
3. Add the results together:
[tex]\[ f(-3) = -27 + 9 - 3 - 2 \][/tex]
[tex]\[ = -27 + 9 = -18 \][/tex]
[tex]\[ -18 - 3 = -21 \][/tex]
[tex]\[ -21 - 2 = -23 \][/tex]
Therefore, the remainder when [tex]\( x^3 + x^2 + x - 2 \)[/tex] is divided by [tex]\( x + 3 \)[/tex] is [tex]\(-23\)[/tex].
### 3b). If [tex]\(x-1\)[/tex] is a factor of [tex]\(x^3 - 7x + 6\)[/tex], what is the remainder when [tex]\(x^3 - 7x + 6\)[/tex] is divided by [tex]\(x-1\)[/tex]?
Given that [tex]\(x-1\)[/tex] is a factor of [tex]\(x^3 - 7x + 6\)[/tex], it means that when we divide [tex]\( f(x) = x^3 - 7x + 6 \)[/tex] by [tex]\( x - 1 \)[/tex], the remainder should be [tex]\( 0 \)[/tex]. This is derived from the Factor Theorem, which states that if [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( f(x) \)[/tex], then [tex]\( f(a) = 0 \)[/tex].
1. Confirm that [tex]\(x = 1\)[/tex] makes the polynomial zero:
[tex]\[ f(1) = 1^3 - 7(1) + 6 \][/tex]
2. Simplify:
[tex]\[ f(1) = 1 - 7 + 6 \][/tex]
[tex]\[ = 1 - 7 = -6 \][/tex]
[tex]\[ -6 + 6 = 0 \][/tex]
Therefore, the remainder when [tex]\( x^3 - 7x + 6 \)[/tex] is divided by [tex]\( x - 1 \)[/tex] is [tex]\(0\)[/tex].
### 3a). Find the remainder when [tex]\(x^3 + x^2 + x - 2\)[/tex] is divided by [tex]\(x + 3\)[/tex]
We will use the Remainder Theorem for this. The Remainder Theorem states that the remainder of the division of a polynomial [tex]\( f(x) \)[/tex] by a linear divisor [tex]\( x - a \)[/tex] is [tex]\( f(a) \)[/tex].
In this case, we are dividing [tex]\( f(x) = x^3 + x^2 + x - 2 \)[/tex] by [tex]\( x + 3 \)[/tex]. Notice that [tex]\( x + 3 \)[/tex] can be written as [tex]\( x - (-3) \)[/tex]. Hence, [tex]\( a = -3 \)[/tex].
1. Substitute [tex]\( x = -3 \)[/tex] into the polynomial:
[tex]\[ f(-3) = (-3)^3 + (-3)^2 + (-3) - 2 \][/tex]
2. Calculate each term:
[tex]\[ (-3)^3 = -27 \][/tex]
[tex]\[ (-3)^2 = 9 \][/tex]
[tex]\[ (-3) = -3 \][/tex]
[tex]\[ -2 = -2 \][/tex]
3. Add the results together:
[tex]\[ f(-3) = -27 + 9 - 3 - 2 \][/tex]
[tex]\[ = -27 + 9 = -18 \][/tex]
[tex]\[ -18 - 3 = -21 \][/tex]
[tex]\[ -21 - 2 = -23 \][/tex]
Therefore, the remainder when [tex]\( x^3 + x^2 + x - 2 \)[/tex] is divided by [tex]\( x + 3 \)[/tex] is [tex]\(-23\)[/tex].
### 3b). If [tex]\(x-1\)[/tex] is a factor of [tex]\(x^3 - 7x + 6\)[/tex], what is the remainder when [tex]\(x^3 - 7x + 6\)[/tex] is divided by [tex]\(x-1\)[/tex]?
Given that [tex]\(x-1\)[/tex] is a factor of [tex]\(x^3 - 7x + 6\)[/tex], it means that when we divide [tex]\( f(x) = x^3 - 7x + 6 \)[/tex] by [tex]\( x - 1 \)[/tex], the remainder should be [tex]\( 0 \)[/tex]. This is derived from the Factor Theorem, which states that if [tex]\( x - a \)[/tex] is a factor of a polynomial [tex]\( f(x) \)[/tex], then [tex]\( f(a) = 0 \)[/tex].
1. Confirm that [tex]\(x = 1\)[/tex] makes the polynomial zero:
[tex]\[ f(1) = 1^3 - 7(1) + 6 \][/tex]
2. Simplify:
[tex]\[ f(1) = 1 - 7 + 6 \][/tex]
[tex]\[ = 1 - 7 = -6 \][/tex]
[tex]\[ -6 + 6 = 0 \][/tex]
Therefore, the remainder when [tex]\( x^3 - 7x + 6 \)[/tex] is divided by [tex]\( x - 1 \)[/tex] is [tex]\(0\)[/tex].
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