How much is y⁴ - 12y² + y + 14 greater than 17y³ + 34y² - 51y + 68
Solution:
We have to determine by how much is y⁴ - 12y² + y + 14 greater than 17y³ + 34y² - 51y + 68.
Required expression = y⁴ - 12y² + y + 14 - (17y³ + 34y² - 51y + 68)
= y⁴ - 12y² + y + 14 - 17y³ - 34y² + 51y - 68
By combining like terms,
= y⁴ - 12y² - 34y² - 17y³ + y + 51y + 14 - 68
= y⁴ - 17y³ + (-12 - 34)y² + (1 + 51)y + (14 - 68)
= y⁴ - 17y³ + (-46)y² + (52)y + (-54)
= y⁴ - 17y³ - 46y² + 52y - 54
Therefore, the required expression is y⁴ - 17y³ - 46y² + 52y - 54
✦ Try This: How much is 2y⁴ - 14y² + 3y + 15 greater than 19y³ + 39y² - 53y + 78
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 12
NCERT Exemplar Class 7 Maths Chapter 10 Problem 62
How much is y⁴ - 12y² + y + 14 greater than 17y³ + 34y² - 51y + 68
Summary:
y⁴ - 12y² + y + 14 is greater than 17y³ + 34y² - 51y + 68 by y⁴ - 17y³ - 46y² + 52y - 54
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