Question
Using factor theorem, factorize the following polynomials:
y3 - 7y + 6
y3 - 7y + 6
✓
Answer
Let f(y) = y3 - 7y + 6
The factors of constant term in f(y) are $\pm1,\pm,2,\pm3$ and $\pm6.$
We have,
f(1) = 1 - 7 + 6 = 0
⇒ (y - 1) is a factor of f(y)
f(-1) = -1 + 7 + 6 = 12
⇒ (y + 1) is a factor of f(y)
f(2) = 8 - 14 + 6 = 0
⇒ (y - 2) is a factor of f(y)
f(-2) = -8 + 14 + 6 = 12
⇒ (y + 2) is not a factor of f(y)
f(3) = 27 - 21 + 6 = 12
⇒ (y - 3) is not a factor of f(y)
f(-3) = -27 + 21 + 6 = 0
⇒ (y + 3) is a factor of f(y)
Since f(y) is a polynomial of degree 3. So, it cannot have more than 3 linear factors.
Thus, factors of f(y) are (y - 1)(y - 2) and (y + 3).
Therefore,
f(y) = k(y - 1)(y - 2)(y + 3)
y3 - 7y + 6 = k(y - 1)(y - 2)(y + 3) ...(1)
Putting y = 0 on both sides, we get,
6 = k(-1)(-2)(3)
6 = 6k
k = 1
Substituting k = 1 in (1), we get,
y3 - 7y + 6 = (y - 1)(y - 2)(y + 3)
The factors of constant term in f(y) are $\pm1,\pm,2,\pm3$ and $\pm6.$
We have,
f(1) = 1 - 7 + 6 = 0
⇒ (y - 1) is a factor of f(y)
f(-1) = -1 + 7 + 6 = 12
⇒ (y + 1) is a factor of f(y)
f(2) = 8 - 14 + 6 = 0
⇒ (y - 2) is a factor of f(y)
f(-2) = -8 + 14 + 6 = 12
⇒ (y + 2) is not a factor of f(y)
f(3) = 27 - 21 + 6 = 12
⇒ (y - 3) is not a factor of f(y)
f(-3) = -27 + 21 + 6 = 0
⇒ (y + 3) is a factor of f(y)
Since f(y) is a polynomial of degree 3. So, it cannot have more than 3 linear factors.
Thus, factors of f(y) are (y - 1)(y - 2) and (y + 3).
Therefore,
f(y) = k(y - 1)(y - 2)(y + 3)
y3 - 7y + 6 = k(y - 1)(y - 2)(y + 3) ...(1)
Putting y = 0 on both sides, we get,
6 = k(-1)(-2)(3)
6 = 6k
k = 1
Substituting k = 1 in (1), we get,
y3 - 7y + 6 = (y - 1)(y - 2)(y + 3)
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