Question

solve this # irrelavent answers will be reported.​

Answer 1

Explanation:

Solution:

Let p(x)=x3–2x2–x+2

Factors of 2 are ±1 and ± 2

By trial method, we find that

p(1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x3–2x2–x+2

p(−1)=(−1)3–2(−1)2–(−1)+2

=−1−2+1+2

=0

Therefore, (x+1) is the factor of p(x)

ncert solutions for class 9 maths chapter 1 fig 1

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2–3x+2) =(x+1)(x2–x–2x+2)

=(x+1)(x(x−1)−2(x−1))

=(x+1)(x−1)(x-2)

(ii) x3–3x2–9x–5

Solution:

Let p(x) = x3–3x2–9x–5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x3–3x2–9x–5

p(5) = (5)3–3(5)2–9(5)–5

=125−75−45−5

=0

Therefore, (x-5) is the factor of p(x)

ncert solutions for class 9 maths chapter 1 fig 2

Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x2+2x+1) =(x−5)(x2+x+x+1)

=(x−5)(x(x+1)+1(x+1))

=(x−5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

Let p(x) = x3+13x2+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x) = x3+13x2+32x+20

p(-1) = (−1)3+13(−1)2+32(−1)+20

=−1+13−32+20

=0

Therefore, (x+1) is the factor of p(x)

ncert solutions for class 9 maths chapter 1 fig 3

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2+12x+20) =(x+1)(x2+2x+10x+20)

=(x+1)x(x+2)+10(x+2)

solve this # irrelavent answers will be reported

=(x+1)(x+2)(x+10)

Answer 2

Answer:

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