Show that this quadratic factor is positive for all real value of x

Question

Let $p(x) = 4x^3 - 4x^2 + 5x + 4$, how can I show that the quadratic factor of $p(x)$ is positive for all real value of x ?

I already found the factor of p(x) and it is $(2x+1)$ and $2x^2-3x+4$, but I'm not sure what to do next to show that $2x^2-3x+4$ is positive for all real value of x.

Answer 1

If you have found $(2x+1)$ as a factor, then we have $$4x^3-4x^2+5x+4 = (2x+1)(ax^2+bx+c)$$ Expanding $(2x+1)(ax^2+bx+c)$, we get $$2ax^3 + (a+2b)x^2+(b+2c)x+c = 4x^3-4x^2+5x+4$$ Comparing coefficients, we get that \begin{align} 2a & = 4\\ a+2b & = -4\\ b+2c & = 5\\ c & = 4 \end{align} Hence, we get the quadratic factor as $$(2x^2-3x+4)$$ Now note that the quadratic factor can be written as follows. \begin{align} 2x^2-3x+4 & = 2\left(x^2 - \dfrac32 x + 2 \right) = 2\left(x^2 - 2 \cdot x \cdot \dfrac34 + \left(\dfrac34 \right)^2 - \left(\dfrac34 \right)^2 + 2 \right)\\ & = 2 \left(\underbrace{\left(x-\dfrac34\right)^2}_{\text{Square is non-negative}} + \dfrac{23}{16}\right) \geq \dfrac{23}8 \end{align} and hence is strictly non-negative.

Answer 2

$p'(x)=12x^2-8x+5=12(x-\frac{1}{3})^2+\frac{16}{3}>0$, so $p(x)$ intersects $x$ at at most one point. Consider that the leading coefficient of the quadratic factor is positive. If the quadratic factor is not always positive then $p(x)$ should intersect $x$ at three points, a contradiction.

Answer 3

Following from user-17762’s result, the quadratic factor is

$$(2x^2-3x+4) = f(x), say$$

$((x^2)) = 2 > 0$ implies --- the graph of $f(x)$ concave (opening) upwards ………. (1)

$⊿ = (-3)^2 – 4(2)(4) < 0$ implies --- the graph of $y = f(x)$ will never cut the x-axis …….. (2)

(1) & (2) together implies the $y$ of $f(x) > 0$ for all real $x$.