If x^3+8x^2+kx+18 is completely divisible by x^2+6x+9,then find the value of k
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Answered by
74
x^3 + 8x^2 + kx + 18 is completely divisible by x^2 + 6x + 9
first of all we should break x^2 + 6x + 9
e.g., x^2 + 2.3x + 3^2 = (x + 3)^2
hence, x = -3 is the zero of x^3 + 8x^2 + kx + 18
now, put x = -3 in x^3 + 8x^2 + kx + 18
(-3)^3 + 8(-3)^2 + k(-3) + 18 = 0
⇒ -27 + 8 × 9 -3k + 18 =0
⇒ -27 + 72 - 3k + 18 = 0
⇒ 63 - 3k = 0
⇒ k = 63/3 = 21
hence, k = 21
first of all we should break x^2 + 6x + 9
e.g., x^2 + 2.3x + 3^2 = (x + 3)^2
hence, x = -3 is the zero of x^3 + 8x^2 + kx + 18
now, put x = -3 in x^3 + 8x^2 + kx + 18
(-3)^3 + 8(-3)^2 + k(-3) + 18 = 0
⇒ -27 + 8 × 9 -3k + 18 =0
⇒ -27 + 72 - 3k + 18 = 0
⇒ 63 - 3k = 0
⇒ k = 63/3 = 21
hence, k = 21
Answered by
112
Given ⇒
f(x) = x³ + 8x² + kx = 18
Divisor of the Polynomial = x² + 6x + 9
∴ x² + 6x + 9 = 0
Splitting the middle term,
x² + 3x + 3x + 9 = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0
(x + 3)² = 0
∴ x + 3 = 0
∴ x = -3
Now, (x + 3) can divide the polynomial f(x) completely.
∴ Remainder f(x) = 0
∴ 0 = (-3)³ + 8(-3)² + k(-3) + 18
∴ 0 = -27 + 8(9) -3k + 18
∴ 72 - 27 - 3k + 18 = 0
⇒ 3k = 45 + 18
⇒ k = 63/3
∴ k = 21
Hence, the value of the k is 21.
Hope it helps.
f(x) = x³ + 8x² + kx = 18
Divisor of the Polynomial = x² + 6x + 9
∴ x² + 6x + 9 = 0
Splitting the middle term,
x² + 3x + 3x + 9 = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0
(x + 3)² = 0
∴ x + 3 = 0
∴ x = -3
Now, (x + 3) can divide the polynomial f(x) completely.
∴ Remainder f(x) = 0
∴ 0 = (-3)³ + 8(-3)² + k(-3) + 18
∴ 0 = -27 + 8(9) -3k + 18
∴ 72 - 27 - 3k + 18 = 0
⇒ 3k = 45 + 18
⇒ k = 63/3
∴ k = 21
Hence, the value of the k is 21.
Hope it helps.
sheebaazadazad:
Here, why are we splitting the middle term in the divisor?
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