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Sure, let's prove that [tex]\(\{P(x) + Q(x)\} - \{R(x) + S(x)\} \)[/tex] is a zero polynomial, meaning the result is always zero for any value of [tex]\(x\)[/tex]:
Given:
[tex]\[ P(x) = 2x^2(x + 3) - (7x - 10) \][/tex]
[tex]\[ Q(x) = 8(x - 1)(x^2 + x + 1) - 3x^2 + 4 \][/tex]
[tex]\[ R(x) = (2x + 1)^2 - (2x^2 + 7) \][/tex]
[tex]\[ S(x) = 2(5x^3 + 6) + x(x - 11) \][/tex]
First, let's simplify each polynomial.
### Simplify [tex]\(P(x)\)[/tex]:
[tex]\[ P(x) = 2x^2(x + 3) - (7x - 10) \][/tex]
[tex]\[ = 2x^2 \cdot x + 2x^2 \cdot 3 - 7x + 10 \][/tex]
[tex]\[ = 2x^3 + 6x^2 - 7x + 10 \][/tex]
### Simplify [tex]\(Q(x)\)[/tex]:
[tex]\[ Q(x) = 8(x - 1)(x^2 + x + 1) - 3x^2 + 4 \][/tex]
Expanding [tex]\(8(x - 1)(x^2 + x + 1)\)[/tex]:
[tex]\[ = 8 \left( x^3 + x^2 + x - x^2 - x - 1 \right) \][/tex]
[tex]\[ = 8 \left( x^3 - 1 \right) \][/tex]
[tex]\[ = 8x^3 - 8 \][/tex]
Adding the remaining terms:
[tex]\[ Q(x) = 8x^3 - 8 - 3x^2 + 4 \][/tex]
[tex]\[ = 8x^3 - 3x^2 - 4 \][/tex]
### Simplify [tex]\(R(x)\)[/tex]:
[tex]\[ R(x) = (2x + 1)^2 - (2x^2 + 7) \][/tex]
Expanding [tex]\((2x + 1)^2\)[/tex]:
[tex]\[ = 4x^2 + 4x + 1 \][/tex]
[tex]\[ R(x) = 4x^2 + 4x + 1 - 2x^2 - 7 \][/tex]
[tex]\[ = 2x^2 + 4x - 6 \][/tex]
### Simplify [tex]\(S(x)\)[/tex]:
[tex]\[ S(x) = 2(5x^3 + 6) + x(x - 11) \][/tex]
[tex]\[ = 10x^3 + 12 + x^2 - 11x \][/tex]
[tex]\[ = 10x^3 + x^2 - 11x + 12 \][/tex]
Now, let's find [tex]\(P(x) + Q(x)\)[/tex] and [tex]\(R(x) + S(x)\)[/tex]:
### Calculate [tex]\(P(x) + Q(x)\)[/tex]:
[tex]\[ P(x) + Q(x) = (2x^3 + 6x^2 - 7x + 10) + (8x^3 - 3x^2 - 4) \][/tex]
[tex]\[ = 2x^3 + 8x^3 + 6x^2 - 3x^2 - 7x - 4 + 10 \][/tex]
[tex]\[ = 10x^3 + 3x^2 - 7x + 6 \][/tex]
### Calculate [tex]\(R(x) + S(x)\)[/tex]:
[tex]\[ R(x) + S(x) = (2x^2 + 4x - 6) + (10x^3 + x^2 - 11x + 12) \][/tex]
[tex]\[ = 10x^3 + 2x^2 + x^2 + 4x - 11x - 6 + 12 \][/tex]
[tex]\[ = 10x^3 + 3x^2 - 7x + 6 \][/tex]
Observe that [tex]\(P(x) + Q(x) = R(x) + S(x)\)[/tex]:
[tex]\[ P(x) + Q(x) = 10x^3 + 3x^2 - 7x + 6 \][/tex]
[tex]\[ R(x) + S(x) = 10x^3 + 3x^2 - 7x + 6 \][/tex]
Hence, [tex]\(\{P(x) + Q(x)\} - \{R(x) + S(x)\} = (10x^3 + 3x^2 - 7x + 6) - (10x^3 + 3x^2 - 7x + 6) \)[/tex]:
[tex]\[ \{P(x) + Q(x)\} - \{R(x) + S(x)\} = 0 \][/tex]
Therefore, [tex]\(\{P(x) + Q(x)\} - \{R(x) + S(x)\} \)[/tex] is indeed a zero polynomial, proving the given statement.
Given:
[tex]\[ P(x) = 2x^2(x + 3) - (7x - 10) \][/tex]
[tex]\[ Q(x) = 8(x - 1)(x^2 + x + 1) - 3x^2 + 4 \][/tex]
[tex]\[ R(x) = (2x + 1)^2 - (2x^2 + 7) \][/tex]
[tex]\[ S(x) = 2(5x^3 + 6) + x(x - 11) \][/tex]
First, let's simplify each polynomial.
### Simplify [tex]\(P(x)\)[/tex]:
[tex]\[ P(x) = 2x^2(x + 3) - (7x - 10) \][/tex]
[tex]\[ = 2x^2 \cdot x + 2x^2 \cdot 3 - 7x + 10 \][/tex]
[tex]\[ = 2x^3 + 6x^2 - 7x + 10 \][/tex]
### Simplify [tex]\(Q(x)\)[/tex]:
[tex]\[ Q(x) = 8(x - 1)(x^2 + x + 1) - 3x^2 + 4 \][/tex]
Expanding [tex]\(8(x - 1)(x^2 + x + 1)\)[/tex]:
[tex]\[ = 8 \left( x^3 + x^2 + x - x^2 - x - 1 \right) \][/tex]
[tex]\[ = 8 \left( x^3 - 1 \right) \][/tex]
[tex]\[ = 8x^3 - 8 \][/tex]
Adding the remaining terms:
[tex]\[ Q(x) = 8x^3 - 8 - 3x^2 + 4 \][/tex]
[tex]\[ = 8x^3 - 3x^2 - 4 \][/tex]
### Simplify [tex]\(R(x)\)[/tex]:
[tex]\[ R(x) = (2x + 1)^2 - (2x^2 + 7) \][/tex]
Expanding [tex]\((2x + 1)^2\)[/tex]:
[tex]\[ = 4x^2 + 4x + 1 \][/tex]
[tex]\[ R(x) = 4x^2 + 4x + 1 - 2x^2 - 7 \][/tex]
[tex]\[ = 2x^2 + 4x - 6 \][/tex]
### Simplify [tex]\(S(x)\)[/tex]:
[tex]\[ S(x) = 2(5x^3 + 6) + x(x - 11) \][/tex]
[tex]\[ = 10x^3 + 12 + x^2 - 11x \][/tex]
[tex]\[ = 10x^3 + x^2 - 11x + 12 \][/tex]
Now, let's find [tex]\(P(x) + Q(x)\)[/tex] and [tex]\(R(x) + S(x)\)[/tex]:
### Calculate [tex]\(P(x) + Q(x)\)[/tex]:
[tex]\[ P(x) + Q(x) = (2x^3 + 6x^2 - 7x + 10) + (8x^3 - 3x^2 - 4) \][/tex]
[tex]\[ = 2x^3 + 8x^3 + 6x^2 - 3x^2 - 7x - 4 + 10 \][/tex]
[tex]\[ = 10x^3 + 3x^2 - 7x + 6 \][/tex]
### Calculate [tex]\(R(x) + S(x)\)[/tex]:
[tex]\[ R(x) + S(x) = (2x^2 + 4x - 6) + (10x^3 + x^2 - 11x + 12) \][/tex]
[tex]\[ = 10x^3 + 2x^2 + x^2 + 4x - 11x - 6 + 12 \][/tex]
[tex]\[ = 10x^3 + 3x^2 - 7x + 6 \][/tex]
Observe that [tex]\(P(x) + Q(x) = R(x) + S(x)\)[/tex]:
[tex]\[ P(x) + Q(x) = 10x^3 + 3x^2 - 7x + 6 \][/tex]
[tex]\[ R(x) + S(x) = 10x^3 + 3x^2 - 7x + 6 \][/tex]
Hence, [tex]\(\{P(x) + Q(x)\} - \{R(x) + S(x)\} = (10x^3 + 3x^2 - 7x + 6) - (10x^3 + 3x^2 - 7x + 6) \)[/tex]:
[tex]\[ \{P(x) + Q(x)\} - \{R(x) + S(x)\} = 0 \][/tex]
Therefore, [tex]\(\{P(x) + Q(x)\} - \{R(x) + S(x)\} \)[/tex] is indeed a zero polynomial, proving the given statement.
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