To determine which expression is not a factor of the polynomial \( 6x^3 - 35x^2 - 69x + 140 \), we will use polynomial long division for each expression one by one. We seek to find which division leaves a remainder.
Let's perform the long division for each expression:
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Test \( x + 5 \) Divide \( 6x^3 - 35x^2 - 69x + 140 \) by \( x + 5 \):
\[ 6x^3 \div x = 6x^2 \] Multiply \( 6x^2 \) by \( x + 5 \): \[ 6x^2 \cdot (x + 5) = 6x^3 + 30x^2 \] Subtract from the original polynomial: \[ (6x^3 - 35x^2 - 69x + 140) - (6x^3 + 30x^2) = -65x^2 - 69x + 140 \]
Divide \( -65x^2 \) by \( x \): \[ -65x^2 \div x = -65x \] Multiply \( -65x \) by \( x + 5 \): \[ -65x \cdot (x + 5) = -65x^2 - 325x \] Subtract: \[ (-65x^2 - 69x + 140) - (-65x^2 - 325x) = 256x + 140 \]
Divide \( 256x \) by \( x \): \[ 256x \div x = 256 \] Multiply \( 256 \) by \( x + 5 \): \[ 256 \cdot (x + 5) = 256x + 1280 \] Subtract: \[ (256x + 140) - (256x + 1280) = -1140 \]
Since the remainder is \(-1140\), \( x + 5 \) is NOT a factor.
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Test \( x - 7 \) Divide \( 6x^3 - 35x^2 - 69x + 140 \) by \( x - 7 \):
\[ 6x^3 \div x = 6x^2 \] Multiply \( 6x^2 \) by \( x - 7 \): \[ 6x^2 \cdot (x - 7) = 6x^3 - 42x^2 \] Subtract: \[ (6x^3 - 35x^2 - 69x + 140) - (6x^3 - 42x^2) = 7x^2 - 69x + 140 \]
Divide \( 7x^2 \) by \( x \): \[ 7x^2 \div x = 7x \] Multiply \( 7x \) by \( x - 7 \): \[ 7x \cdot (x - 7) = 7x^2 - 49x \] Subtract: \[ (7x^2 - 69x + 140) - (7x^2 - 49x) = -20x + 140 \]
Divide \( -20x \) by \( x \): \[ -20x \div x = -20 \] Multiply: \[ -20 \cdot (x - 7) = -20x + 140 \] Subtract: \[ (-20x + 140) - (-20x + 140) = 0 \]
Since the remainder is \(0\), \( x - 7 \) is a factor.
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Test \( 2x + 5 \) Divide \( 6x^3 - 35x^2 - 69x + 140 \) by \( 2x + 5 \):
\[ 6x^3 \div 2x = 3x^2 \] Multiply \( 3x^2 \) by \( 2x + 5 \): \[ 3x^2 \cdot (2x + 5) = 6x^3 + 15x^2 \] Subtract: \[ (6x^3 - 35x^2 - 69x + 140) - (6x^3 + 15x^2) = -50x^2 - 69x + 140 \]
Divide \( -50x^2 \) by \( 2x \): \[ -50x^2 \div 2x = -25x \] Multiply: \[ -25x \cdot (2x + 5) = -50x^2 - 125x \] Subtract: \[ (-50x^2 - 69x + 140) - (-50x^2 - 125x) = 56x + 140 \]
Divide \( 56x \) by \( 2x \): \[ 56x \div 2x = 28 \] Multiply: \[ 28 \cdot (2x + 5) = 56x + 140 \] Subtract: \[ (56x + 140) - (56x + 140) = 0 \]
Since the remainder is \(0\), \( 2x + 5 \) is a factor.
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Test \( 3x - 4 \) Divide \( 6x^3 - 35x^2 - 69x + 140 \) by \( 3x - 4 \):
\[ 6x^3 \div 3x = 2x^2 \] Multiply: \[ 2x^2 \cdot (3x - 4) = 6x^3 - 8x^2 \] Subtract: \[ (6x^3 - 35x^2 - 69x + 140) - (6x^3 - 8x^2) = -27x^2 - 69x + 140 \]
Divide \( -27x^2 \) by \( 3x \): \[ -27x^2 \div 3x = -9x \] Multiply: \[ -9x \cdot (3x - 4) = -27x^2 + 36x \] Subtract: \[ (-27x^2 - 69x + 140) - (-27x^2 + 36x) = -105x + 140 \]
Divide \( -105x \) by \( 3x \): \[ -105x \div 3x = -35 \] Multiply: \[ -35 \cdot (3x - 4) = -105x + 140 \] Subtract: \[ (-105x + 140) - (-105x + 140) = 0 \]
Since the remainder is \(0\), \( 3x - 4 \) is also a factor.
Based on the checks, the expression \( x + 5 \) is not a factor of \( 6x^3 - 35x^2 - 69x + 140 \).
Answer: \( x + 5 \)