factor this expression: $4x^4-49x^2+56x-16$

Question

I am having a hard time factoring this expression. please explain to me the process of factoring this. my classmate arrived at the answer $(x+4)(2x-1)(2x^2-7x+4)$. how do i arrive at the same answer? thank you!

Answer 1

Write

$$4x^4-49x^2+56x-16 = (ax^2 + bx + c)(\alpha x^2 + \beta x + \gamma)$$ Then multiply out the brackets and note that \begin{align} a \alpha &= 4 \\ b \beta +ac +\alpha\gamma &= -49 \\ &\vdots \end{align} And try to solve for each coefficient noting that the coefficient in $x^3$ is zero.

Answer 2

HINT: after the rational root theorem we get $$x_1=-4$$ and $$x_2=\frac{1}{2}$$ are solutions

Answer 3

You can let $f(x)$ be the expression and apply the Factor Theorem.

i.e. If $f(a)=0$, then $(x-a)$ is a factor.

Identify two factors first then perform long division or comparing coefficients to find the quadratic factor.

Answer 4

Hint:

Observe that $\;49x^2-56x+16=(7x-4)^2$.

Answer 5

$4x^4-49x^2+56x-16=4x^4-\left(49x^2-56x+16\right)=\left(2x^2\right)^2-\left(7x-4\right)^2=\left(2x^2-7x+4\right)\left(2x^2+7x-4\right).$

Or you should try to verify if the integer divisors of the free member ($-16$) are the solutions of this expression.

Answer 6

For all real $k$ we have $$4x^4-49x^2+56x-16=(2x^2+k)^2-4kx^2-k^2-49x^2+56x-16=$$ $$=(2x^2+k)^2-((4k+49)x^2-56x+k^2+16).$$ Now, we'll choose $k$ such that $(4k+49)x^2-56x+k^2+16=(ax+b)^2$, for which we need $$28^2-(4k+49)(k^2+16)=0.$$ Easy to see that $k=0$ is valid.

Thus, $$4x^4-49x^2+56x-16=(2x^2)^2-(49x^2-56x+16)=$$ $$=(2x^2)^2-(7x-4)^2=(2x^2-7x+4)(2x^2+7x-4)$$ and the rest is smooth.

Answer 7

Although 'tricks' may be found to factor a $4^{th}$-degree polynomial, in general it is not an easy thing to do.

But
$\tag 1 4x^4-49x^2+56x-16$

is special in that the coefficients are all integers, so you can immediately apply the Rational root theorem as sketched by Dr. Sonnhard Graubner. So any rational root must be among the numbers symbolically indicated by,

$\tag 2 {\displaystyle \pm \left\{1,2,4,8,16,{\frac {1}{2}},{\frac {1}{4}}\right\}}$

Now you can plug all these number into (1) looking for a root, but to speed things up just plug your polynomial into Wolfram's factoring calculator. When you do, it you get rational solutions you are on your way. You can organize your work by dividing out your 'chosen' linear factor and then continue looking for rational roots in the $3^{rd}$-degree polynomial factor.

For (1) you get two linear factors with rational roots, but using the quadratic formula you can factor "all the way" using square roots.