Question

16. Find the value of p and q, if (x + 3) and (x - 4) are
factors of x^3-px^2-qx+24
​

Answer 1

Hey mate....

Here is your answer...

Answer 2

Answer:

Concept:

Polynomials and Factorization of polynomials.

Step-by-step explanation:

Given:

$\ f(x)=x^{3}+a x^{2}-b x+24$

Find:

Values of p and q

Solution:

Let us assume

$x+3=0 \\Then, x=-3 \\ Now, substitute \ the \ value \ of \ x \ in \ \mathrm{f}(\mathrm{x}) \begin{aligned}&f(-3)=(-3)^{3}+a(-3)^{2}-b(-3)+24 \\&=-27+9 a+3 b+24 \\&=9 a+3 b-3\end{aligned} \\Dividing \ all \ terms \ by \ 3 \ we \ get, =3 a+b-1 \\From \ the \ question, \\ (x+3) \ is \ a \ factor \ of \ x^{3}+a x^{2}-b x+24 .\\Therefore, remainder is 0 . \begin{aligned}&f(x)=0 \\&3 a+b-1=0 \\&3 a+b=1 \ldots \text { [equation (i)] }\end{aligned}$

$Now, \\assume \ x-4=0 \\Then,\\ x=4 \\Given, f(x)=x^{3}+a x^{2}-b x+24 \\Now, substitute the value of x in \mathrm{f}(\mathrm{x}) , \begin{aligned}&f(4)=43+a(4)^{2}-b(4)+24 \\&=64+16 a-4 b+24 \\&=88+16 a-4 b\end{aligned} Dividing all terms by 4 we get, =22+4 a-b \begin{aligned}&\mathrm{f}(\mathrm{x})=0 \\&22+4 \mathrm{a}-\mathrm{b}=0 \\&4 \mathrm{a}-\mathrm{b}=-22 \ldots \text { [equation (ii)] }\end{aligned}$

Now, add the two equations (i) and (ii)

$\begin{aligned}&(3 a+b)+(4 a-b)=1-22 \\&3 a+b+4 a-b=-21 \\&7 a=-21 \\&a=-21 / 7 \\&a=-3\end{aligned} Consider \ the \ equation (i) to \ find \ out 'b'. \begin{aligned}&3 a+b=1 \\&3(-3)+b=1 \\&-9+b=1 \\&b=1+9 \\&b=10\end{aligned} Therefore, value \ of a=-3 and b=10 .\\Then, \ by \ substituting \ the \ value \ of \ a \ and \ b\\ f(x)=x^{3}-3 x^{2}-10 x+24$

$\begin{aligned}&(x+3)(x-4) \\&=x(x-4)+3(x-4) \\&=x 2-4 x+3 x-12 \\&=x^{2}-x-12\end{aligned} Dividing f(x) by x^{2}-x-12 we get,\\Therefore, x^{3}-3 x 2-10 x+24=(x 2-x-12)(x-2) =(x+3)(x-4)(x-2)$

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