Question

solve these :

1] The HCF of polynomials x⁴+6x²+25 ; x³-3x²+7x-5 ; and x²+5-2x is?

2]The HCF of the polynomials (2x-1)(5x²-ax+3) and (x-3)(2x²+x+b) is (2x-1)(x-3) , then find the value of a and b ? ​

Answer 1

Required Answer:-

31) Factorising the higher degree polynomials first.

x⁴ + 6x² + 25

(x²)² + 2.x².5 + (5)² - 4x²

(x² + 5)² - (2x)²

Using identity (a² - b²) = (a + b)(a - b)

(x² + 5 - 2x)(x² + 5 + 2x)

x³ - 3x² + 7x - 5

Split in such a way to get a common.

x³ - x² - 2x² + 2x + 5x - 5

x²(x - 1) - 2x(x - 1) + 5(x - 1)

(x² - 2x + 5)(x - 1)

And the third polynomial is x² - 2x + 5. Here we can see that x² - 2x + 5 is common in all the three polynomials and if we factories x² - 2x + 5, no common factor will be the highest (as for HCF). Hence, the HCF of polynomials is x² - 2x + 5 Option B

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32) HCF of (2x - 1)(5x² - ax + 3) and (x - 3)(2x² + x + b) is (2x - 1)(x - 3).

In the first polynomial 2x - 1 is already a factor, so (5x² - ax + 3) must be divisible by x - 3. Then p(3) = 0 by using factor theoram$.$ Putting x = 3

5(3)² - a(3) + 3 = 0

45 - 3a + 3 = 0

48 - 3a = 0

a = 16

In the second polynomial x - 3 is already a factor, so (2x² + x + b) must be divisible by 2x - 1. Then p(1/2) = 0 by using factor theoram. Putting x = 1/2

2(1/2)² + 1/2 + b = 0

2(1/4) + 1/2 + b = 0

1 + b = 0

b = -1

Thus, the correct option is 16, -1 (A)

Answer 2

Required Answer:-

31) Factorising the higher degree polynomials first.

x⁴ + 6x² + 25

(x²)² + 2.x².5 + (5)² - 4x²

(x² + 5)² - (2x)²

Using identity (a² - b²) = (a + b)(a - b)

(x² + 5 - 2x)(x² + 5 + 2x)

x³ - 3x² + 7x - 5

Split in such a way to get a common.

x³ - x² - 2x² + 2x + 5x - 5

x²(x - 1) - 2x(x - 1) + 5(x - 1)

(x² - 2x + 5)(x - 1)

And the third polynomial is x² - 2x + 5. Here we can see that x² - 2x + 5 is common in all the three polynomials and if we factories x² - 2x + 5, no common factor will be the highest (as for HCF). Hence, the HCF of polynomials is x² - 2x + 5 Option B

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32) HCF of (2x - 1)(5x² - ax + 3) and (x - 3)(2x² + x + b) is (2x - 1)(x - 3).

In the first polynomial 2x - 1 is already a factor, so (5x² - ax + 3) must be divisible by x - 3. Then p(3) = 0 by using factor theoram.. Putting x = 3

5(3)² - a(3) + 3 = 0

45 - 3a + 3 = 0

48 - 3a = 0

a = 16

In the second polynomial x - 3 is already a factor, so (2x² + x + b) must be divisible by 2x - 1. Then p(1/2) = 0 by using factor theoram. Putting x = 1/2

2(1/2)² + 1/2 + b = 0

2(1/4) + 1/2 + b = 0

1 + b = 0

b = -1

Thus, the correct option is 16, -1 (A