Answer:
Explanation:
This "subtraction problem" involves: "one value" minus "another value":
The first value is: " 10m³n²σ⁵ " ;
The second value is: 25m⁴nσ " ;
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The problem to be solved is:
" 10m³n²σ⁵ − 25m⁴nσ = ?? ;
Let us "factor out" what we can—considering the two (2) values:
1) " 10m³n²σ⁵ " ; and:
2) " 25m⁴nσ " ;
The GCF ["Greatest Common Factor"] of:
"10" and "25" :
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List the factors for Both "10" and for "25"—as follows:
10: 1, 2, 5, 10.
25: 1, 5, 25.
→ The GCF for "10 and 25" is "5".
→ Note that the GCF for " m³ and m⁴ " is "m³ ."
→ Note that the GCF for: " n² and n" is "n". {Note: " n = n¹ ".}
→ Note that the GCF for: " σ⁵ and "σ" is "σ". {Note: " σ = σ¹ ".}
So, the GCF for entire problem:
Factor out: " 5m³nσ"
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The first expression:
10m³n²σ⁵ ;
Factor out: (5m³nσ} ;
(5m³nσ} (2*1*nσ⁴}
= (5m³nσ} (2nσ⁴}
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The second expression:
25m⁴nσ ;
Factor out:
(5m³nσ} (5m*1*1} ;
= {5m³nσ} (5m} .
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So:
(5m³nσ} (2nσ⁴} − (5m³nσ} (5m} ;
= {5m³nσ} * (2nσ⁴ − 5m) ; {Note: {" ax − ay = a(x − y) "}.
Now, simplify:
{5m³nσ} * (2nσ⁴ − 5m) =
{5m³nσ} * 1(2nσ⁴ − 5m) ;
→ 1 (2nσ⁴ − 5m) ; Note: a(b+c) = ab + ac ;
a(b−c) = ab − ac ;
1 (2nσ⁴ − 5m) = 2nσ⁴ + (-5m)
= 2nσ⁴ − 5m ;
Now bring down the:
{5m³nσ} ; and rewrite:
⇒ {5m³nσ}*(2nσ⁴ − 5m); Note: a(b+c) = ab + ac ; ==> again:
in which: a = 5m³nσ ; b = 2nσ⁴ ; c = -5m ;
→ {5m³nσ}*(2nσ⁴ - 5m} =
[5m³nσ}*(2nσ⁴} + {5m³nσ}{-5m) ;
Start with the "left-hand side" of the problem; ^directly above:
[5m³nσ}*(2no⁴} = 5*2*m³*n²σ⁵ = 10m³*n²σ⁵ ;
Then: We shall examine the "right-hand side" of the above:
"{5m³nσ}{-5m)" ;
→ {-25m⁴nσ);
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Then, we shall add these values together:
10m³*n²σ⁵ + (-25m⁴nσ);
= 10m³*n²σ⁵ − 25m⁴nσ);