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Question
factor ( x^4 - 8x^2 + 16 ) completely. all factors in your answer should have integer coefficients.
Step1: Let \( y = x^2 \), rewrite the expression.
The given expression is \( x^4 - 8x^2 + 16 \). Let \( y=x^{2}\), then the expression becomes \( y^{2}-8y + 16 \).
Step2: Factor the quadratic in \( y \).
We know that for a quadratic \( ay^{2}+by + c \), if \( a = 1\), \( b=-8\), \( c = 16\), we can factor it as \( (y - m)(y - n) \) where \( m + n=b\) and \( mn=c\). Here, \( m=n = 4\) since \( 4+4=8\) and \( 4\times4 = 16\). So \( y^{2}-8y + 16=(y - 4)^{2}\).
Step3: Substitute back \( y=x^{2} \).
Substitute \( y = x^{2}\) into \( (y - 4)^{2}\), we get \( (x^{2}-4)^{2} \).
Step4: Factor \( x^{2}-4 \) further.
Using the difference of squares formula \( a^{2}-b^{2}=(a - b)(a + b) \), where \( a=x\) and \( b = 2\), so \( x^{2}-4=(x - 2)(x + 2) \).
Step5: Write the final factored form.
Then \( (x^{2}-4)^{2}=[(x - 2)(x + 2)]^{2}=(x - 2)^{2}(x + 2)^{2} \).
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\((x - 2)^{2}(x + 2)^{2}\)