Question

if y^4 -3y^2+7y -10 is divided by (y-2)​

Answer 1

Answer:

y⁴-3y²+7y-10 is divided by (y-2)

so, y-2

y-2=0

y=2........

(now put the value)

y⁴-3y²+7y-10

2⁴-3(2)²+7(2)-10

16-3(4)+7(2)-10

16-12+14-10

30-22

8 not equal to 0

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Answer 2

Value = 8

GIVEN :-

p(y) =

${y}^{4} - {3y}^{2} + 7y - 10$

divided by

$(y - 2)$

To Find:- Value

SOLUTION:-

According to remainder theorem when a polynomial is divided by a linear polynomial then the remainder is equal to zero.

Therefore,

$y - 2 = 0$

$y = 2$

By substituting the value of p in the given polynomial p(y)

${y}^{4 } - {3y}^{2} + 7y - 10$

$= {2}^{4} - (3 \times {2}^{2} ) + (7 \times 2) - 10$

$= 16 - (3 \times 4) + 14 - 10$

$= 16 - 12 + 14 - 10$

$= 30 - 22$

$= 8$

Hence, value = 8

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