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Factorize :

$(i) 2x^{3}-3x^{2}-17x+30$

$(ii) x^{3}-6x^{2}+11x-6$

$(iii) x^{3}+x^{2}-4x-4$

$(iv) 3x^{3}-x^{2}-3x+1$

Answers (1)

$(i) (x-2)(x+3)(2x-5)$

Solution

Let $p(x)= 2x^{3}-3x^{2}-17x+30$

By trail, we find that $p(2)=0$

$p(2)= 2(2)^{3}-3(2)^{2}-17(2)+30=16-12-34+30=0$

Hence $x-2$ is a factor of $p(x)$

$\\2x^{3}-3x^{2}-17x+30=\left ( x-2 \right )\left ( 2x^{2}+x-15 \right )\\ =(x-2)(2x^{2}+6x-5x-15)\\ =(x-2)(2x(x+3)-5(x+3))\\ 2x^{3}-3x^{2}-17x+30= (x-2)(x+3)(2x-5)$

Hence the factorized form is $(x-2)(x+3)(2x-5)$

$(ii) (x-1)(x-2)(x-3)$

Solution

Let $p(x)= x^{3}-6x^{2}+11x-6$

By trail, we find that $p(1)=0$

$p(1)=1-6+11-6=0$

Hence $x-1$ is a factor of $p(x)$

$\\p(x)=(x-1)(x^{2}-5x+6)\\ =(x-1)(x^{2}-3x-2x+6))\\ =(x-1)(x(x-3)-2(x-3))\\= (x-1)(x-2)(x-3)$

Hence the factorized form is $(x-1)(x-2)(x-3)$

$(iii) (x+1)(x-2)(x+2)$

Solution

Let $p(x)=x^{3}+x^{2}-4x-4$

By trail, we find that $p(-1)=0$

$p(-1)=-1+1+4-4=0$

Hence $x+1$ is a factor of $p(x)$

$\\p(x)=(x+1)(x^{2}-4)\\ =(x+1)(x^{2}-(2)^{2}))\\ =(x+1)(x-2)(x+2)\; \; \; \; (using a^{2}-b^{2}=(a-b)(a+b))\\= (x+1)(x-2)(x+2)$

Hence the factorized form is $(x+1)(x-2)(x+2)$

$(iv) (x-1)(x+1)(3x-1)$

Solution

Let $p(x)=3x^{3}-x^{2}-3x-1$

By trail, we find that $p(1)=0$

$\\p(1)=3(1)^{3}-(1)^{2}+3(1)+1\\=3-1-3+1=0$

Hence $x-1$ is a factor of $p(x)$

$\\p(x)=(x-1)(3x^{3}+2x-1)\\ =(x-1)(3x^{2}+3x-x-1)\\ =(x-1)(3x(x+1)-1(x+1)) \\= (x-1)(x+1)(3x-1)$

Hence the factorized form is $(x-1)(x+1)(3x-1)$ .

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