Question

The H.C.F and L.C.M of two polynomials of the third degree are x² + x+5 and x⁴ + 2x³+ 4x² + 3x-10 respectively. Find the two polynomials.

Answer 1

Given:

H.C.F. = $x^{2}+x+5$

L.C.M. = $x^{4}+2x^{3}+4x^{2}+3x-10$

To find:

The two polynomials of third degree

Step-by-step explanation:

We know that, the product of the two polynomial

= H.C.F. × L.C.M.

= $(x^{2}+x+5)(x^{4}+2x^{3}+4x^{2}+3x-10)$

= $x^{6}+2x^{5}+4x^{4}+3x^{3}-10x^{2}+x^{5}+2x^{4}+4x^{3}+3x^{2}-10x+5x^{4}+10x^{3}+20x^{2}+15x-50$

= $x^{6}+3x^{5}+11x^{4}+17x^{3}+13x^{2}+5x-50$

= $x^{6}-x^{5}+4x^{5}-4x^{4}+15x^{4}-15x^{3}+32x^{3}-32x^{2}+45x^{2}-45x+50x-50$

= $x^{5}(x-1)+4x^{4}(x-1)+15x^{3}(x-1)+32x^{2}(x-1)+45x(x-1)+50(x-1)$

= $(x-1)(x^{5}+4x^{4}+15x^{3}+32x^{2}+45x+50)$

= $(x-1)(x^{5}+2x^{4}+2x^{4}+4x^{3}+11x^{3}+22x^{2}+10x^{2}+20x+25x+50)$

= $(x-1)\{x^{4}(x+2)+2x^{3}(x+2)+11x^{2}(x+2)+10x(x+2)+25(x+2)\}$

= $(x-1)(x+2)(x^{4}+2x^{3}+11x^{2}+10x+25)$

= $(x-1)(x+2)(x^{4}+x^{3}+5x^{2}+x^{3}+x^{2}+5x+5x^{2}+5x+25)$

= $(x-1)(x+2)\{x^{2}(x^{2}+x+5)+x(x^{2}+x+5)+5(x^{2}+x+5)\}$

= $(x-1)(x+2)(x^{2}+x+5)(x^{2}+x+5)$

= $(x-1)(x^{2}+x+5)(x+2)(x^{2}+x+5)$

= $(x^{3}+x^{2}+5x-x^{2}-x-5)(x^{3}+x^{2}+5x+2x^{2}+2x+10)$

= $(x^{3}+4x-5)(x^{3}+3x^{2}+7x+10)$

Answer:

Thus the two required polynomials are $(x^{3}+4x-5)$ and $(x^{3}+3x^{2}+7x+10)$.

Note:

• In order to solve the given problem, firstly we have multiplied H.C.F. and L.C.M. given.
• Then we have applied root finding method to factorize the product.
• Finally we have rearranged the factors to find polynomials of the third degree.