Question

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$(x - 2)$ is a factor of $x^4 + 2x^3 - 7x^2 - 8x + 12$. the other factors are , , and .
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Explanation:

Step1: Perform Polynomial Division or Use Synthetic Division

Since \((x - 2)\) is a factor, we can use polynomial long division or synthetic division to divide \(x^{4}+2x^{3}-7x^{2}-8x + 12\) by \((x - 2)\). Let's use synthetic division for \(x=2\):

The coefficients of the polynomial are \(1,2,-7,-8,12\).

Bring down the first coefficient: \(1\)

Multiply by \(2\): \(1\times2 = 2\), add to next coefficient: \(2 + 2=4\)

Multiply by \(2\): \(4\times2 = 8\), add to next coefficient: \(-7+8 = 1\)

Multiply by \(2\): \(1\times2 = 2\), add to next coefficient: \(-8 + 2=-6\)

Multiply by \(2\): \(-6\times2=-12\), add to last coefficient: \(12-12 = 0\) (which confirms divisibility)

So the quotient is \(x^{3}+4x^{2}+x - 6\)

Step2: Factor the Cubic Polynomial \(x^{3}+4x^{2}+x - 6\)

Try to find a root of the cubic. Let's test \(x = 1\): \(1^{3}+4\times1^{2}+1-6=1 + 4+1 - 6=0\). So \((x - 1)\) is a factor.

Now divide \(x^{3}+4x^{2}+x - 6\) by \((x - 1)\) using synthetic division for \(x = 1\):

Coefficients: \(1,4,1,-6\)

Bring down \(1\), multiply by \(1\): \(1\), add to \(4\): \(5\)

Multiply by \(1\): \(5\), add to \(1\): \(6\)

Multiply by \(1\): \(6\), add to \(-6\): \(0\)

Quotient is \(x^{2}+5x + 6\)

Step3: Factor the Quadratic Polynomial \(x^{2}+5x + 6\)

Factor \(x^{2}+5x + 6\). We need two numbers that multiply to \(6\) and add to \(5\). Those numbers are \(2\) and \(3\). So \(x^{2}+5x + 6=(x + 2)(x + 3)\)

Step4: Combine All Factors

The original polynomial \(x^{4}+2x^{3}-7x^{2}-8x + 12=(x - 2)(x - 1)(x + 2)(x + 3)\)

So the other factors (besides \((x - 2)\)) are \((x - 1)\), \((x + 2)\), and \((x + 3)\)

Answer:

The other factors are \((x - 1)\), \((x + 2)\), and \((x + 3)\)