Answer:
Please check below the step-by-step explanation.
Step-by-step explanation:
Considering the quadratic expression
![ax^2+bx\:+\:c\:=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%5C%3A%2B%5C%3Ac%5C%3A%3D0)
![a,\:b,\:and\:c\:can\:have\:any\:value,\:except\:that\:a\:can't\:be\:0.](https://tex.z-dn.net/?f=a%2C%5C%3Ab%2C%5C%3Aand%5C%3Ac%5C%3Acan%5C%3Ahave%5C%3Aany%5C%3Avalue%2C%5C%3Aexcept%5C%3Athat%5C%3Aa%5C%3Acan%27t%5C%3Abe%5C%3A0.)
![To\:Factorise\:a\:Quadratic\:is\:to:](https://tex.z-dn.net/?f=To%5C%3AFactorise%5C%3Aa%5C%3AQuadratic%5C%3Ais%5C%3Ato%3A)
"
"
For example, multiplying
and
together will bring
i.e.
![\left(x+3\right)\left(x-2\right) = xx+x\left(-2\right)+3x+3\left(-2\right)](https://tex.z-dn.net/?f=%5Cleft%28x%2B3%5Cright%29%5Cleft%28x-2%5Cright%29%20%3D%20xx%2Bx%5Cleft%28-2%5Cright%29%2B3x%2B3%5Cleft%28-2%5Cright%29)
![\mathrm{Apply\:minus-plus\:rules}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aminus-plus%5C%3Arules%7D)
![+\left(-a\right)=-a](https://tex.z-dn.net/?f=%2B%5Cleft%28-a%5Cright%29%3D-a)
![\left(x+3\right)\left(x-2\right) = xx-2x+3x-3\cdot \:2](https://tex.z-dn.net/?f=%5Cleft%28x%2B3%5Cright%29%5Cleft%28x-2%5Cright%29%20%3D%20xx-2x%2B3x-3%5Ccdot%20%5C%3A2)
![= x^2+x-6](https://tex.z-dn.net/?f=%3D%20x%5E2%2Bx-6)
Therefore,
and
are factors of ![x^2+x-6](https://tex.z-dn.net/?f=x%5E2%2Bx-6)
Keywords: factorize, quadratic expression
Learn more about quadratic expression from brainly.com/question/12182022
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