Question 1 Report
If (x + 2) and (x - 1) are factors of \(f(x) = 6x^{4} + mx^{3} - 13x^{2} + nx + 14\), find the
(a) values of m and n.
(b) remainder when f(x) is divided be (x + 1).
Given \(f(x)=6x^4+mx^3-13x^2+nx+14,\) with \((x+2)\) and \((x-1)\) as factors, so \(f(-2)=0\) and \(f(1)=0.\)
(a) Values of m and n.
\(f(-2)=6(16)+m(-8)-13(4)+n(-2)+14=96-8m-52-2n+14=58-8m-2n=0.\)
So \(8m+2n=58\Rightarrow 4m+n=29.\quad(1)\)
\(f(1)=6+m-13+n+14=7+m+n=0\Rightarrow m+n=-7.\quad(2)\)
Subtract (2) from (1): \(3m=36\Rightarrow m=12.\) Then \(n=-7-12=-19.\)
\[m=12,\qquad n=-19.\]
(b) Remainder when f(x) is divided by \((x+1).\)
By the remainder theorem the remainder is \(f(-1).\) With \(f(x)=6x^4+12x^3-13x^2-19x+14:\)
\[f(-1)=6(1)+12(-1)-13(1)-19(-1)+14=6-12-13+19+14=14.\]
Remainder \(=14.\)
Answer Details
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