Answer:
x=2 ; x=-4 or x=-3/2
Explanation:
We are given a cubic polynomial
Here as we are having a cubic equation to factorize, we have to find some value of x such that when we put that value in place of x , it gives us a 0. We have to try it randomly like first we try it for x=1 then x=2 and so on.
On trying for x=2 , we see that the equation becomes 0=0 and hence x=2 is one of the solution. And also as x=2 is one of the solution, (x-2) must be one of the factor. Now we use this property to determine the other factor like this.
![2x^3+7x^2-10x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/h3devrhuomdwnk1dhsakwk7or5w54d4vdt.png)
Adding and subtracing
to
we get
![2x^3-4x^2+4x^2+7x^2-10x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4a6ufq8yb3tf98ct0rutunj9y28qsl0859.png)
Now we take out
as GCF
![2x^2(x-2)+11x^2-10x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/84tn4m60l1ya35sa4sh198nn2q9ackajk9.png)
Now subtracting and adding
to
![11x^2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/cumkon97huh9hzkj6mpzq4gkwmzllc58ly.png)
![2x^2(x-2)+11x^2-22x+22x-10x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2sgww4y1zwsmd41yfndqhyt27ba6rj2mn0.png)
taking
as GCF out
![2x^2(x-2)+11x(x-2)+22x-10x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gb5vu62uu0frafm2ale93ioodze4uj8k5z.png)
![2x^2(x-2)+11x(x-2)+12x-24=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gn788bo794nbwtbsljzt0hggxqgz84herw.png)
![2x^2(x-2)+11x(x-2)+12(x-2)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/1yhr5s24umsgbo4imuvjn1xm5vvziya3r7.png)
Now taking
out as GCF
![(x-2)(2x^2+11x+12)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t31s8osmrac3ywk0bcavmypk8f0gbk6hbn.png)
Now we need to split the middle term of
to factorize it in such a way that their product is 2*12 and sum is 11. We have 8 and 3 as these factors.
Hence it can be factored like this
![(2x^2+11x+12)\\2x^2+8x+3x+12\\2x(x+4)+3(x+4)\\(2x+3)(x+4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/n7tf5sudfeoof1iopqg2p2n49fb92dmwha.png)
Hence our answer will be
![(x-2)(2x+3)(x+4)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/552bdvddngbc6qwh603tn53dsalwxss2a2.png)
And thus the solutions will be
x-2=0 : x=2
x+4=0 ; x=-4
2x+3=0 ; x=-
![(3)/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/85tww783zdtzzps1k74cyql3cl3s1y42ha.png)