What should be added to x cube + 5 x square + 7 x + 3 so that it is completely divisible by x square + 2 x?
Answers
Answered by
34
Answer:
-x - 3
Step-by-step explanation:
What should be added to x cube + 5 x square + 7 x + 3 so that it is completely divisible by x square + 2 x
x³ + 5x² + 7x + 3
Let say bx + c is added to it make it completely divisible by x² + 2x
divisible by x² + 2x
x ( x + 2)
x = 0 & x = -2 are factor
x³ + 5x² + 7x + 3 + bx + c = 0 for x = 0 & x = -2
=> 0 + 0 + 0 + 3 + 0 + c = 0
=> c = -3
-8 + 20 -14 + 3 -2b -3 = 0
=> -2b = 2
=> b = -1
bx + c = -x - 3
-x -3 need to be added to x³ + 5x² + 7x + 3 to make it completely divisble
by adding this equation will become
x³ + 5x² + 6x
= x(x² + 5x + 6)
= x(x+3)(x+2)
= (x²+2x)(x+3)
Answered by
8
Answer:
-x - 3
Step-by-step explanation:
What should be added to x³ + 5x² + 7x + 3 so that it is completely divisible by x ² + 2 x
x³ + 5x² + 7x + 3
Let say bx + c is added to it make it completely divisible by x² + 2x
divisible by x² + 2x
x ( x + 2)
x = 0 & x = -2 are factor
x³ + 5x² + 7x + 3 + bx + c = 0 for x = 0 & x = -2
⇒ 0 + 0 + 0 + 3 + 0 + c = 0
⇒ c = -3
-8 + 20 -14 + 3 -2b -3 = 0
⇒-2b = 2
⇒ b = -1
bx + c = -x - 3
-x -3 need to be added to x³ + 5x² + 7x + 3 to make it completely divisble
by adding this equation will become
x³ + 5x² + 6x
= x(x² + 5x + 6)
= x(x+3)(x+2)
= (x²+2x)(x+3)
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