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# Factorization Algebra Tricky Question

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Factor $$-16x^4+x^2+2x+1$$  into two quadratic polynomials with integer coefficients. Submit your answer in the form $$(ax^2+bx+c)(dx^2+ex+f)$$ , with $$a . Aug 13, 2022 ### 1+0 Answers #1 +1 Expand the right hand side: \((-16x^4+x^2+2x+1)=adx^4+aex^3+afx^2+bdx^3+bex^2+bfx+cdx^2+cex+cf$$

$$\hspace{4.4cm}=adx^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf$$

So:

$$ad=-16 \\ ae+bd=0 \\ af+be+cd=1 \\ bf+ce=2 \\ cf=1 \\$$

The possible combinations of a and d: $$(a,d)=(4,-4),(16,-1),(8,-2)$$

$$c=1,f=1$$ is a must by the last equation.

So our system becomes:

$$ad=-16 \\ ae+bd=0 \\ a+be+d = 1 \\ b+e = 2 \\$$

It is good idea to try one of the combinations of the above.

Suppose (a,d) = (4,-4)
Then the first equation is satisfied. The second equation is 4e-4b = 0, which only works if e-b=0 or e=b, but from the last equation, b+e = 2 , hence 2e=2 meaning e=1 and thus b=1.

so we solved the system.

Therefore: $$-16x^4+x^2+2x+1=(4x^2+x+1)(-4x^2+x+1)$$

I hope this helps :)!

Aug 13, 2022