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प्रश्न
Show that x + 3 is a factor of 69 + 11x – x2 + x3.
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उत्तर
Let p(x) = x3 – x2 + 11x + 69
We have to show that, x + 3 is a factor of p(x).
i.e., p(–3) = 0
Now, p(–3) = (–3)3 – (–3)2 + 11(–3) + 69
= –27 – 9 – 33 + 69
= – 69 + 69
= 0
Hence, (x + 3) is a factor of p(x).
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संबंधित प्रश्न
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case:
p(x) = x3 − 4x2 + x + 6, g(x) = x − 3
Factorise:
12x2 – 7x + 1
Factorise:
3x2 – x – 4
Factorize the following polynomial.
(y2 + 5y) (y2 + 5y – 2) – 24
Show that 2x – 3 is a factor of x + 2x3 – 9x2 + 12.
Show that p – 1 is a factor of p10 – 1 and also of p11 – 1.
Factorise:
x2 + 9x + 18
Factorise the following:
9x2 – 12x + 3
Factorise:
`2sqrt(2)a^3 + 8b^3 - 27c^3 + 18sqrt(2)abc`
Without finding the cubes, factorise:
(x – 2y)3 + (2y – 3z)3 + (3z – x)3
