answer: our required factorization will be
(3x+4)(x²-5)
step-by-step explanation:
since we have given that
![3x^3+4x^2-15x-20](/tex.php?f=3x^3+4x^2-15x-20)
now, we will write it as :
![=\left(3x^3+4x^2\right)+\left(-15x-20\right)](/tex.php?f==\left(3x^3+4x^2\right)+\left(-15x-20\right))
![\mathrm{factor\: out\: }-5\mathrm{\: from\: }-15x-20\mathrm{: \quad }-5\left(3x+4\right)](/tex.php?f=\mathrm{factor\: out\: }-5\mathrm{\: from\: }-15x-20\mathrm{: \quad }-5\left(3x+4\right))
so, we get
5(3x+4)
similarly,
![\mathrm{factor\: out\: }x^2\mathrm{\: from\: }3x^3+4x^2\mathrm{: \quad }x^2\left(3x+4\right)](/tex.php?f=\mathrm{factor\: out\: }x^2\mathrm{\: from\: }3x^3+4x^2\mathrm{: \quad }x^2\left(3x+4\right))
here, we get,
2(3x+4)
together it becomes,
![-5\left(3x+4\right)+x^2\left(3x+4\right)\\\\=\left(3x+4\right)\left(x^2-5\right)](/tex.php?f=-5\left(3x+4\right)+x^2\left(3x+4\right)\\\\=\left(3x+4\right)\left(x^2-5\right))
hence, our required factorization will be
(3x+4)(x²-5)