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Answered by
1
Answer:
(2−3y−5y
2
) by (2y−1+3y
2
)
= (2−3y−5y
2
)×(2y−1+3y
2
)
= 2(2y−1+3y
2
)−3y(2y−1+3y
2
)−5y
2
(2y−1+3y
2
)
= 4y−2+6y
2
−6y
2
+3y−9y
3
−10y
3
+5y
2
−15y
4
= −15y
4
—19y
3
+5y
2
+7y−2
Answered by
2
Answer:
Hi!
Step-by-step explanation:
2−3y−5y 2) by (2y−1+3y 2) = (2−3y−5y 2)×(2y−1+3y 2) = 2(2y−1+3y 2)−3y(2y−1+3y 2)−5y 2(2y−1+3y 2.
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