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To XXXX the solution to XXXX XXXXXXX, just XXXXXX subtract the XXXXX XXXX XX the XXXX XXXX XXX combined XXX/or total area XX XXXX the lawn XXX shed.
The XXXXX XXXX of XXX XXXX is XXXXX by XXXXXXXXXXX the length XXX width of XXX rectangle defining the XXXX. The length XX XXX shed XX defined by the binomial, (2x + X), and the width of the XXXX XX defined XX XXX XXXXXXXX XXXX, 2x. XXXX, the XXXXXXXXXXXX property is XXXX XX XXXXXXXX the XXXXXXXX XXX XXXXXXXX terms as XXXXXXX:
XXX XXXXXXXX XXX/or XXXXX XXXX XX XXXX the lawn XXX XXXX is found XX XXXXXXXXXXX XXX XXXXXX and width XX the XXXXXXXXX defining the XXXX. XXX length of XXXX XXXXXXXX XXXX is defined by the XXXXXXXX, (XXX− 4), XXX the width of the XXXX XX XXXXXXX by XXX monomial term, XX. Note, XXX XXXXXXXXXXXX property XX XXXX XX multiply the monomial XXX XXXXXXXX XXXXX
XX XXXXXXX:
XXXX, take the XXX areas XXX XXXXXXXX XXX XXXXX XXXX of the shed XXXX XXX combined and/or total area of both the XXXX and shed XX XXXXXXX:
Thus, the polynomial that describes XXX lawn XXXX needs to XXX XX XXXXXXX as XXXXXXX:
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