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 $\alpha, \beta$ are the roots of the equation $x^2 - 2x + 4 = 0$,then the value of $\alpha^n + \beta^n$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{25} x + \sqrt{16} y = 41$
$II.$ $\sqrt{16} x + \sqrt{25} y = 40$

If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

If the graph of $y = ax^2 + bx + c$ $(a, b, c \in R)$ is as shown in the figure,where $D = b^2 - 4ac$,which of the following is incorrect?

Roots of $ax^2 + b = 0$ are real and distinct if

Vedclass Test Series

Exam Paper Generator

Online Exam Module