Factorise 49 x^{2}-42 x+9. | vedclass

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

Question

(B) To factorise the expression $49 x^{2}-42 x+9$,we observe that it follows the algebraic identity $a^{2}-2ab+b^{2} = (a-b)^{2}$.Here,$a^{2} = 49 x^{

Vedclass · Accepted Answer

$(7x-3)(7x-3)$

Vedclass · Answer

(B) To factorise the expression $49 x^{2}-42 x+9$,we observe that it follows the algebraic identity $a^{2}-2ab+b^{2} = (a-b)^{2}$.Here,$a^{2} = 49 x^{2} = (7 x)^{2}$,so $a = 7 x$.And $b^{2} = 9 = (3)^{2}$,so $b = 3$.Now,check the middle term: $-2ab = -2(7 x)(3) = -42 x$.Since the expression matches the identity,we can write:$49 x^{2}-42 x+9 = (7 x)^{2}-2(7 x)(3)+(3)^{2} = (7 x-3)^{2}$.Thus,the factors are $(7 x-3)(7 x-3)$.

Factorise $49 x^{2}-42 x+9$.

Similar Questions

With the help of the remainder theorem,find the remainder when the polynomial $p(x) = x^{3} + x^{2} - 26x + 24$ is divided by the divisor $x - 1$.

Factorise the following expression: $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$

Find $p(1)$,$p(2)$,and $p(4)$ for the following polynomial: $p(y) = y^{2} - 5y + 4$.

Evaluate $(215)^{2}$.

For the polynomial $p(x),$ if $p(7)=0,$ then $\ldots$ is a factor of $p(x).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module