Final challenge: p(x) = 2x^3 + 3x^2 - 11x - 6. A student finds the remainders on division by (x - 2), (x + 3) and (x + 1/2) and notices a pattern. How many of these three divisors leave remainder 0 (i.e. are factors of p(x))?

Question

Accepted Answer

3. p(2) = 16 + 12 - 22 - 6 = 0. p(-3) = -54 + 27 + 33 - 6 = 0. p(-1/2) = 2(-1/8) + 3(1/4) - 11(-1/2) - 6 = -1/4 + 3/4 + 11/2 - 6 = 1/2 + 11/2 - 6 = 6 - 6 = 0. All three give remainder 0.

Final challenge: $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ .
A student finds the remainders on division by $(x - 2), (x + 3)$ and $(x + \frac{1}{2})$ and notices a pattern.
How many of these three divisors leave remainder 0 (i.e. are factors of $p (x)$ )?

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More Class 9 Remainder Theorem practice

Final challenge: p(x)=2x3+3x2−11x−6.A student finds the remainders on division by (x−2),(x+3) and (x+21​) and notices a pattern.How many of these three divisors leave remainder 0 (i.e. are factors of p(x))?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 9 Remainder Theorem practice

Final challenge: $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 6$ .
A student finds the remainders on division by $(x - 2), (x + 3)$ and $(x + \frac{1}{2})$ and notices a pattern.
How many of these three divisors leave remainder 0 (i.e. are factors of $p (x)$ )?