How do you factor the trinomial $49x^2 − 8x + 16$ ?

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Answer 1 · 2015-12-22T22:59:11

You use the formula : $(a-b)^2 = a^2-2ab+b^2$

$a = 7x$
$b = -4/7$

$a^2=49x^2$
$b^2= 16/49$
$2ab = -8x$

$49x^2-8x+16/49-16/49+16$

$(7x-4/7)^2+768/49$

Answer 2 · 2015-12-23T00:03:25

$49x^2-8x+16$

$= (x-4/49-(16sqrt(3)i)/49)(x-4/49+(16sqrt(3)i)/49)$

$49x^2-8x+16$ is of the form $ax^2+bx+c$ with $a=49$ , $b=-8$ and $c=16$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-8)^2-(4xx49xx16) = 64-3136 = -3072$

$= -3*2^10$

Since this is negative, our trinomial has no linear factors with Real coefficients.

It has Complex zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (8+-sqrt(Delta))/98$

$=(8+-32sqrt(3)i)/98 = (4+-16sqrt(3)i)/49$

Hence:

$49x^2-8x+16$

$= (x-4/49-(16sqrt(3)i)/49)(x-4/49+(16sqrt(3)i)/49)$

How do you factor the trinomial 49x^2 − 8x + 16 49x^2 − 8x + 16?