To decompose the polynomial \( (x^2 - 7x + 6)(x^2 + 3x - 18) \) into linear factors, we will first factor each quadratic polynomial separately.
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Factor the first polynomial: \( x^2 - 7x + 6 \)
We need to find two numbers that multiply to \( 6 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)). The numbers \( -1 \) and \( -6 \) satisfy these conditions.
Thus, we can factor it as: \[ x^2 - 7x + 6 = (x - 1)(x - 6) \]
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Factor the second polynomial: \( x^2 + 3x - 18 \)
Now, we need to find two numbers that multiply to \( -18 \) and add up to \( 3 \). The numbers \( 6 \) and \( -3 \) satisfy these conditions.
Thus, we can factor it as: \[ x^2 + 3x - 18 = (x + 6)(x - 3) \]
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Combine the factors:
Now, we can write the entire polynomial as a product of its linear factors: \[ (x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 1)(x - 6)(x + 6)(x - 3) \]
So, the complete decomposition into linear factors is: \[ (x - 1)(x - 6)(x + 6)(x - 3) \]