The expression x^{4}+7 x^{2}+16 can be factor | vedclass

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

Question

(C) To factor the expression $x^{4}+7 x^{2}+16$,we can complete the square.First,rewrite the expression by adding and subtracting $x^{2}$:$x^{4}+7 x^{

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) To factor the expression $x^{4}+7 x^{2}+16$,we can complete the square.First,rewrite the expression by adding and subtracting $x^{2}$:$x^{4}+7 x^{2}+16 = (x^{4}+8 x^{2}+16) - x^{2}$Recognize that $(x^{4}+8 x^{2}+16)$ is a perfect square trinomial:$(x^{4}+8 x^{2}+16) = (x^{2}+4)^{2}$Now,substitute this back into the expression:$(x^{2}+4)^{2} - x^{2}$Use the difference of squares formula,$a^{2}-b^{2} = (a+b)(a-b)$,where $a = (x^{2}+4)$ and $b = x$:$(x^{2}+4+x)(x^{2}+4-x)$Rearranging the terms,we get:$(x^{2}+x+4)(x^{2}-x+4)$

The expression $x^{4}+7 x^{2}+16$ can be factored as:

Similar Questions

If the roots of the given equation $(\cos p - 1)x^2 + (\cos p)x + \sin p = 0$ are real,then

If $x = 2 + 2^{2/3} + 2^{1/3}$,then $x^3 - 6x^2 + 6x = $

If $\alpha_1 < \alpha_2 < \alpha_3 < \alpha_4 < \alpha_5 < \alpha_6$,then the equation $(x - \alpha_1)(x - \alpha_3)(x - \alpha_5) + 3(x - \alpha_2)(x - \alpha_4)(x - \alpha_6) = 0$ has :-

If the roots of the equation $x^2 + 2mx + m^2 - 2m + 6 = 0$ are equal,then the value of $m$ is:

Find the roots of the equation $\sqrt{3y+1} = \sqrt{y-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module