Find the L.C.M. of the polynomials: 16-4x^{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given polynomials are $p(x) = 16 - 4x^{2}$ and $q(x) = x^{2} + x - 6$.Step $1$: Factorize $p(x)$.$p(x) = 16 - 4x^{2} = 4(4 - x^{2}) = 4(2 - x)(2 +

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given polynomials are $p(x) = 16 - 4x^{2}$ and $q(x) = x^{2} + x - 6$.Step $1$: Factorize $p(x)$.$p(x) = 16 - 4x^{2} = 4(4 - x^{2}) = 4(2 - x)(2 + x) = -4(x - 2)(x + 2)$.Step $2$: Factorize $q(x)$.$q(x) = x^{2} + x - 6 = x^{2} + 3x - 2x - 6 = x(x + 3) - 2(x + 3) = (x + 3)(x - 2)$.Step $3$: Find the $L.C.M.$The $L.C.M.$ is the product of the highest powers of all prime factors present in the expressions.$L.C.M. = -4(x - 2)(x + 2)(x + 3)$.Simplifying the expression:$L.C.M. = -4(x^{2} - 4)(x + 3)$.

Find the $L.C.M.$ of the polynomials: $16-4x^{2}$ and $x^{2}+x-6$.

Similar Questions

Find the $G.C.D.$ of $3+13 x-30 x^{2}$ and $25 x^{2}-30 x+9$.

If $(x-1)$ is the $H.C.F.$ of $(x^{2}-1)$ and $[p x^{2}-q(x+1)]$,then:

Find the $G.C.D.$ of the polynomials $(x-2)^{2}(x+3)(x-4)$ and $(x-2)(x+2)(x-5)$.

The $H.C.F.$ (Highest Common Factor) of two polynomials is $(y-7)$ and their $L.C.M.$ is $y^{3}-10y^{2}+11y+70$. If one of the polynomials is $y^{2}-5y-14$,find the other.

Find the $G.C.D.$ of $8(x^{4}-16)$ and $12(x^{3}-8)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module