Composite numbers are numbers that are not prime.
<span><span> y2(q-4)-c(q-4)</span> </span>Final result :<span> (q - 4) • (y2 - c)
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Step by step solution :<span>Step 1 :</span><span>Equation at the end of step 1 :</span><span><span> ((y2) • (q - 4)) - c • (q - 4)
</span><span> Step 2 :</span></span><span>Equation at the end of step 2 :</span><span> y2 • (q - 4) - c • (q - 4)
</span><span>Step 3 :</span>Pulling out like terms :
<span> 3.1 </span> Pull out q-4
After pulling out, we are left with :
(q-4) • (<span> y2</span> * 1 +( c * (-1) ))
Trying to factor as a Difference of Squares :
<span> 3.2 </span> Factoring: <span> y2-c</span>
Theory : A difference of two perfect squares, <span> A2 - B2 </span>can be factored into <span> (A+B) • (A-B)
</span>Proof :<span> (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 <span>- AB + AB </span>- B2 =
<span> A2 - B2</span>
</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication.
Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.
Check : <span> y2 </span>is the square of <span> y1 </span>
Check :<span> <span> c1 </span> is not a square !!
</span>Ruling : Binomial can not be factored as the difference of two perfect squares
Final result :<span> (q - 4) • (y2 - c)
</span><span>
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Answer:
3x-x+2=4 :)
Step-by-step explanation:
Hope this helped! :3
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