If x + k is the GCD of x2 − 2x − 15 and x3 + 27, find the value of k.
Answers
Answered by
74
Hey Mate...
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x2 − 2x − 15
= x2 − 5x + 3x − 15 [− 5x + 3x = −2x, − 5x × 3x = −15x2]
= x(x − 5) + 3(x − 5)
= (x − 5) (x + 3)
x3 + 27
= (x)3 + (3)3
= (x + 3) (x2 − 3x + 9) [a3 + b3 = (a + b) (a2 − ab + b2)]
∴GCD of x2 − 2x − 15 and x3 + 27 is x + 3.
It is given that x + k is the G.C.D of x2 − 2x − 15 and x2 + 27.
Comparing x + 3 with x + k, we get k = 3.
Thus, the value of k is 3.
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_______
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x2 − 2x − 15
= x2 − 5x + 3x − 15 [− 5x + 3x = −2x, − 5x × 3x = −15x2]
= x(x − 5) + 3(x − 5)
= (x − 5) (x + 3)
x3 + 27
= (x)3 + (3)3
= (x + 3) (x2 − 3x + 9) [a3 + b3 = (a + b) (a2 − ab + b2)]
∴GCD of x2 − 2x − 15 and x3 + 27 is x + 3.
It is given that x + k is the G.C.D of x2 − 2x − 15 and x2 + 27.
Comparing x + 3 with x + k, we get k = 3.
Thus, the value of k is 3.
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Answered by
65
Holla ^_^
☸ Required Answer is ⬇⬇ ⬇⬇ ⬇
Given Equation,
⭕ ▶ x²-2x-15
By factorisation method,
▶ x² -5x +3x -15
▶ x (x -5 ) +3 (x-5)
▶ (x+3) (x-5) .
⭕ Equation no. 2
▶ x³ + 27
▶ x³ + 3³
▶ (x+3) ( x² -3x +9) .
Therefore, GCD of both equation is (x+3).
Given GCD of x²-2x-15 is x+k .
By Comparing both GCD, we get k =3
✔ Value of k =3 .
Vielen Dank ♥
☸ Required Answer is ⬇⬇ ⬇⬇ ⬇
Given Equation,
⭕ ▶ x²-2x-15
By factorisation method,
▶ x² -5x +3x -15
▶ x (x -5 ) +3 (x-5)
▶ (x+3) (x-5) .
⭕ Equation no. 2
▶ x³ + 27
▶ x³ + 3³
▶ (x+3) ( x² -3x +9) .
Therefore, GCD of both equation is (x+3).
Given GCD of x²-2x-15 is x+k .
By Comparing both GCD, we get k =3
✔ Value of k =3 .
Vielen Dank ♥
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