Question

Please help in answering these questions related to polynomials .

Answer 1

`\sf\\1.\ Solution:\\(a+b+c)^2=9^2\\or,\ a^2+b^2+c^2+2ab+2bc+2ac=81\\or,\ a^2+b^2+c^2+2(ab+bc+ca)=81\\or,\ a^2+b^2+c^2+2(26)=81\\or,\ a^2+b^2+c^2=29`

`\sf\\2.i.\ 9x^2+6x-8\\=9x^2+(12-6)x-8\\=9x^2+12x-6x-8\\=3x(3x+4)-2(3x+4)\\=(3x+4)(3x-2)`

`\sf\\ii.\ 4p^2-17p-21\\= 4p^2-(21-4)p-21\\=4p^2-21p+4p-21\\=p(4p-21)+1(4p-21)\\=(4p-21)(p+1)`

`\sf\\iii.\ 4x^2+20x+25\\=4x^2+(10+10)x+25\\=4x^2+10x+10x+25\\=2x(2x+5)+5(2x+5)\\=(2x+5)(2x+5)=(2x+5)^2\\`

`\sf\\iv.\ x^2+9x+18\\=x^2+(6+3)x+18\\=x^2+6x+3x+18\\=x(x+6)+3(x+6)\\=(x+6)(x+3)`

`\sf\\4.i.\ 8a^3-b^3-12a^2b+6ab^2\\=(2a)^3-b^3-6ab(2a-b)\\=(2a-b)[(2a)^2-2a.b+b^2]-6ab(2a-b)\\=(2a-b)(4a^2-2ab+b^2)-6ab(2a-b)\\=(2a-b)(4a^2-2ab+b^2)`

Here, you may skip the first and second step after adequate amount of practice.

`\sf\\ii.\ 4a^2+b^2+4ab-8a-4b+4\\=(2a)^2+2(2a)b+b^2-4(2a+b)+4\\=(2a+b)^2-4(2a+b)+4\\Let\ 2a+b=x\\Then\ the\ expression=x^2-2x+4=(x-2)^2=(2a+b-2)^2`

`\sf\\5.\ 6x=29^2-23^2\\or,\ 6x=(29+23)(29-23)\\or,\ 6x=52* 6\\or,\ x=52`

Question number 6 and 7 are in the images.

`\sf\\8.i.\ 2x^3-3x^2-17x+30\\=2x^3-4x^2+x^2-2x-15x+30\\=2x^2(x-2)+x(x-2)-15(x-2)\\=(x-2)(2x^2+x-15)\\=(x-2)\{2x^2+6x-5x-15\}\\=(x-2)\{2x(x+3)-5(x+3)\}\\=(x-2)(x+3)(2x-5)`

Here, I used the zero-factor method of factorization. I found that when x = 2, the value of expression is 0. So, I arranged each terms such that they have (x - 2) as the common factor.

`\sf\\ii.\ x^3-6x^2+11x-6\\=x^3-x^2-5x^2-5x-6x-6\\=x^2(x-1)-5x(x-1)-6(x-1)\\=(x-1)(x^2-5x-6)\\=(x-1)(x^2-3x-2x-6)\\=(x-1)\{x(x-3)-2(x-3)\}\\=(x-1)(x-2)(x-3)`

`\sf\\iii.\ x^3+x^2-4x-4\\=x^2(x+1)-4(x+1)\\=(x+1)(x^2-4)\\=(x+1)(x-2)(x+2)`

`\sf\\iv.\ 1+64x^3\\=(1+4x)(1+4x+16x^2)`

`\sf\\v.\ 27y^3-125z^3\\=(3y-5z)(9y^2+15yz+25z^2)`

`\sf\\vi.\ 3x^3-x^2-3x+1\\=x^2(3x-1)-1(3x-1)\\=(3x-1)(x^2-1)\\=(3x-1)(x-1)(x+1)`

`\sf\\9.\ 125a^3+8b^3+c^3-30abc\\=(5a+2b)^3-3(5a)(2b)(5a+2b)+c^3-30abc\\=(5a+2b)^3+c^3-30ab(5a+2b)-30abc\\=(5a+2b+c)\{(5a+2b)^2-(5a+2b)c+c^2\}-30ab(5a+2b)-30abc\\=(5a+2b+c)(25a^2+20ab+4b^2-5ac-2bc+c^2)-30ab(5a+2b+c)\\=(5a+2b+c)(25a^2+20ab+4b^2-5ac-2bc+c^2-30ab)\\=(5a+2b+c)(25a^2-10ab+4b^2-5ac-2bc+c^2)`

`\sf\\10.\ 48^3-30^3-18^3\\=(48-30)^3+3(48)(30)(48-30)-18^3\\=18^3+3(48)(30)(18)-18^3\\=3(48)(30)(18)\\=77760`

`\sf\\11.\ 4x^2+9y^2+16z^2+12xy-24yz-16xz\\=(2x+3y)^2-2(2x)(3y)+16z^2+12xy-24yz-16xz\\=(2x+3y)^2+16z^2-12xy+12xy-24yz-16xz\\=(2x+3y-4z)^2+2(2x+3y)(4z)-12xy+12xy-24yz-16xz\ \ \ \\=(2x+3y-4z)^2+16xz+24yz-12xy+12xy-24yz-16xz\\=(2x+3y-4z)^2`

`\sf\\12.\ 27x^3+64y^3\\=(3x+4y)(9x^2-12xy+16y^2)`

`\sf\\13.i.\ 103^3\\=(100+3)^3\\=100^3+3(100)^2(3)+3(100)(3)^3+3^3\\=10000+90000+8100+27\\=1092727`

`\sf\\(ii)\ 101* 102\\=(100+1)(100+2)\\=100^2+200+100+2\\=10000+300+2\\=10302`

`\sf\\(iii)\ 999^2\\=(1000-1)^2\\=1000^2-2000+1\\=1000000-2000+1\\=998001`

`\sf\\14.\ (x-2y)^3+(2y-3z)^3+(3z-x)^3\\=(x-2y+2y-3z)^3-3(x-2y)(2y-3z)(x-2y+2y-3z)+(3z-x)^3\\=(x-3z)^3-3(x-2y)(2y-3z)(x-3z)-(x-3z)^3\\=-3(x-2y)(2y-3z)(x-3z)`

`\sf\\14.\ 4a^2+4a-3\\=4a^2+6a-2a-3\\=2a(2a+3)-1(2a+3)\\=(2a+3)(2a-1)\ or\ (2a+3)(2a-1)\\\textsf{i.e. either length = (2a+3) and breadth = (2a-1)}\\\textsf{or length = (2a-1) and breadth = (2a+3)}`