Question

the HCF of the expression x square minus 12 x +35 and x square _8x+7​

Answer 1

Answer:

Also, factorizing 2x2 – 3xy by taking the common factor 'x', we get

= x(2x – 3y)

Therefore, H.C.F. of the polynomial 4x2 - 9y2 and 2x2 – 3xy is (2x - 3y).

2. Find the H.C.F. of the polynomials x2 + 4x + 4 and x2 – 4.

Solution:

Factorizing x2 + 4x + 4 by using the identities (a + b)2, we get

(x)2 + 2(x)(2) + (2)2

= (x + 2)2

= (x + 2) (x + 2)

Also, factorizing x2 – 4, we get

(x)2 – (2)2, by using the identities of a2 - b2.

= (x + 2) (x - 2)

Therefore, H.C.F. of x2 + 4x + 4 and x2 – 4 is (x + 2).

3. Find the highest common factor of polynomials x2 + 15x + 56, x2 + 5x - 24 and x2 + 8x.

Solution:

Factorizing x2 + 15x + 56 by splitting the middle term, we get

(x)2 + 8x + 7x + 56

= x(x + 8) + 7(x + 8)

= (x + 8) (x + 7)

Factorizing x2 + 5x - 24, we get

(x)2 + 8x - 3x - 24

= x(x + 8) - 3(x + 8)

= (x + 8) (x - 3)

Factorizing x2 + 8x by taking the common factor 'x', we get

= x(x + 8)

Therefore, H.C.F. of x2 + 15x + 56, x2 + 5x - 24 and x2 + 8x is (x + 8).

4. Find the H.C.F. x2 – 5x + 4, x2 – 2x + 1 and x2 – 1.

Solution:

Factorizing the quadratic trinomial x2 – 5x + 4, we get

(x)2 – x – 4x + 4

= x(x - 1) – 4(x – 1)

= (x - 4) (x - 1)

Factorizing x2 – 2x + 1 by using the identities (a - b)2, we get

(x)2 – 2 (x) (1) + (1)2

= (x – 1)2

Factorizing x2 – 1 by using the differences of two squares, we get

= x2 – 12

= (x + 1) (x – 1)

Therefore, H.C.F. of x2 – 5x + 4, x2 – 2x + 1 and x2 – 1 is (x – 1).