X ^3 + 6x^2 +12x+8 ÷(x+2)
2 answers:
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Hey there!
We can first factor out the 4 (bring it outside the parenthesis)
Now have a closer look between the parenthesis, because this expression can be factored as well. We're looking for two different numbers with a product of -6 and a sum of 5. Hence, we find the numbers 6 and -1
~ Hope this helps you!
We can first factor out the 4 (bring it outside the parenthesis)
Now have a closer look between the parenthesis, because this expression can be factored as well. We're looking for two different numbers with a product of -6 and a sum of 5. Hence, we find the numbers 6 and -1
~ Hope this helps you!
Answer:
Transitive property of equality is not a justification for the proof.
Step-by-step explanation:
We draw a right angle ΔACB. CD is perpendicular to AB.
Let AC = a , BC = b , AB = c and CD = h
Now in ΔABC and ΔACD
∠C = ∠D and ∠A = ∠A
from AA similarity postulate
ΔABC similar to ΔACD.
Hence,
=
= c × x ·····················(1)
Now in ΔABC and ΔCBD
∠C = ∠D and ∠B = ∠B
from AA similarity postulates
ΔABC similar to ΔCBD
Hence,
= c × y······················(2)
Add equation (1) and (2)
+
= cx + cy
+
= c(x+y)
+
=
[because x+y=c]
Transitive property is not useful for this proof.
