The required polynomial is [tex]f(x)=x^{2}-13 x-12=0[/tex]
SOLUTION:
Given,
Zeroes of a polynomial with degree 3 are – 3, - 1, and 4; f( -2 ) = 30
We have to find the polynomial equation.
Now, we know that, general polynomial equation with degree is [tex]\mathrm{z} x^{2}+\mathrm{bx}^{2}+\mathrm{cx}+\mathrm{d}=0[/tex]
So, let [tex]f(x) = 3 x^{2}+b x^{2}+c x+d[/tex]
Now, we know that, [tex]f(-2) = 30 \rightarrow a(-2)^{2}+b(-2)^{2}+c(-2)+d=30 \rightarrow-8 a+4 b-2 c+d=30 \rightarrow(1)[/tex]
Now, -3 is a root, so [tex]f(-3) = 0 \rightarrow a(-3)^{3}+b(-3)^{2}+c(-3)+d=0 \rightarrow-27 a+9 b-3 c+d=0 \rightarrow(2)[/tex]
Now, - 1 is a root, so [tex]f(-1) = 0 \rightarrow a(-1)^{2}+b(-1)^{2}+c(-1)+d=0 \rightarrow-a+b-c+d=0 \rightarrow(3)[/tex]
Now, 4 is a root, so [tex]f(4) = 0 \rightarrow a(4)^{2}+b(4)^{2}+c(4)+d=0 \rightarrow 64 a+16 b+4 c+d=0 \rightarrow(4)[/tex]
So, let us solve the above equations to find value of a, b, c, and d
Now, subtract (3) form (1), (2) and (4)
[tex]\begin{array}{l}{(1) \rightarrow(-8 a+4 b-2 c+d)-(-a+b-8+d)=30-0} \\\\ {\rightarrow-8 a+a+4 b-2 c+c+d-d=0 \rightarrow-7 a+3 b-c=30 \rightarrow(5)} \\\\ {(2) \rightarrow(-27 a+9 b-3 c+d)-(-a+b-c+d=0-30) \rightarrow(5)} \\\\ {\rightarrow-27 a+a+9 b-b-3 c+c+d-d-a \rightarrow-28 a+8 b-2 c=0 \rightarrow(6)} \\\\ {(4) \rightarrow(84 a+16 b+4 c+d)-(-a+b-c+d)=0} \\\\ {\rightarrow 84 a+a+16 b-b+4 c+c+d-d=0 \rightarrow 65 a+15 b+5 c=0 \rightarrow 13 a+3 b+c=0 \rightarrow(7)}\end{array}[/tex]
[tex]\begin{array}{l}{\text { Now, add } 2 \times(7) \text { with }(6) \rightarrow(-26 a+8 b-2 c)+2(13 a+3 b+c)=0} \\\\ {\rightarrow-26 a+8 b-2 c+28 a+6 b+2 c=0 \rightarrow 0+14 b+0=0 \rightarrow 14 b=0 \rightarrow b=0} \\\\ {\text { Now, add }(5) \text { with }(7) \rightarrow(-7 a+3 b-c)+(13 a+3 b+c)=30} \\\\ {\rightarrow 13 a-7 a+3 b+3 b+c-c=30 \rightarrow 6 a+6 b=30 \rightarrow 6 a=30(a=b=0) \rightarrow a=5}\end{array}[/tex]
[tex]\text { Then, put } a=5 \text { and } b=0 \text { in }(5) \rightarrow-7(5)+3(0)-c=30 \rightarrow-35-c=30 \rightarrow c=-35-30 \rightarrow c=-65[/tex]
[tex]\text { And, now, substitute } c=-65 \text { and } a=5 \text { and } b=0 \text { in }(3) \rightarrow-5+0-(-85)+d=0 \rightarrow d+65-5=0 \rightarrow d=-80[/tex]
[tex]\text { Then, } f(x)=5 x^{2}-65 x-60=0 \rightarrow f(x)=x^{2}-13 x-12=0[/tex]
