Advertisements
Advertisements
प्रश्न
How much is y4 – 12y2 + y + 14 greater than 17y3 + 34y2 – 51y + 68?
Advertisements
उत्तर
Required expression is
y4 – 12y2 + y + 14 – (17y3 + 34y2 – 51y + 68)
= y4 – 12y2 + y + 14 – 17y3 – 34y2 + 51y – 68
On combining the like terms,
= y4 – 12y2 – 34y2 + y + 51y + 14 – 68 – 17y3
= y4 – 46y2 + 52y – 17y3 – 54
= y4 – 17y3 – 46y2 + 52y – 54
So, y4 – 12y2 + y + 14 is y4 – 17y3 – 46y2 + 52y – 54 greater than 17y3 + 34y2 – 51y + 68.
APPEARS IN
संबंधित प्रश्न
Simplify combining like terms: - z2 + 13z2 − 5z + 7z3 − 15z
Add: t - 8tz, 3tz - z, z - t
Add the following algebraic expression:
4xy2 − 7x2y, 12x2y − 6xy2, − 3x2y +5xy2
If \[x + \frac{1}{x} = 12,\] find the value of \[x - \frac{1}{x} .\]
Add:
13x2 − 12y2; 6x2 − 8y2
Solve the following equation.
10 = 2y + 5
Add: −9y, 11y, 2y
Add:
3a(a – b + c), 2b(a – b + c)
Add the following expressions:
ab + bc + ca and – bc – ca – ab
What should be added to 3pq + 5p2q2 + p3 to get p3 + 2p2q2 + 4pq?
