Question

factor each of the following expressions.
a $x^{6}+15x^{3}+56=$
b $x^{4}-3x^{2}-28=$
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Explanation:

Step1: Let \( y = x^{2} \)

For part a: \( x^{4}+15x^{2}+56 \) becomes \( y^{2}+15y + 56 \)
For part b: \( x^{4}-3x^{2}-28 \) becomes \( y^{2}-3y - 28 \)

Step2: Factor the quadratic in \( y \)

For part a:
We need two numbers that multiply to \( 56 \) and add up to \( 15 \). The numbers are \( 7 \) and \( 8 \).
So, \( y^{2}+15y + 56=(y + 7)(y + 8) \)
Substituting back \( y=x^{2} \), we get \( (x^{2}+7)(x^{2}+8) \)
For part b:
We need two numbers that multiply to \(- 28\) and add up to \(-3\). The numbers are \( -7 \) and \( 4 \)
So, \( y^{2}-3y - 28=(y - 7)(y+4) \)
Substituting back \( y = x^{2} \), we get \( (x^{2}-7)(x^{2}+4) \)

Answer:

a. \( (x^{2}+7)(x^{2}+8) \)
b. \( (x^{2}-7)(x^{2}+4) \)