Difference between revisions of "1965 AHSME Problems/Problem 19"
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| − | + | == Problem 19== | |
| + | |||
| + | If <math>x^4 + 4x^3 + 6px^2 + 4qx + r</math> is exactly divisible by <math>x^3 + 3x^2 + 9x + 3</math>, the value of <math>(p + q)r</math> is: | ||
| + | |||
| + | <math>\textbf{(A)}\ - 18 \qquad | ||
| + | \textbf{(B) }\ 12 \qquad | ||
| + | \textbf{(C) }\ 15 \qquad | ||
| + | \textbf{(D) }\ 27 \qquad | ||
| + | \textbf{(E) }\ 45 \qquad </math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | Let <math>f(x)=x^3+3x^2+9x+3</math> and <math>g(x)=x^4+4x^3+6px^2+4qx+r</math>. | ||
| + | |||
| + | Let 3 roots of <math>f(x)</math> be <math>r_1, r_2 </math> and <math>r_3</math>. As <math>f(x)|g(x)</math> , 3 roots of 4 roots of <math>g(x)</math> will be same as roots of <math>f(x)</math>. Let the 4th root of <math>g(x)</math> be <math>r_4</math>. By vieta's formula | ||
| + | |||
| + | In <math>f(x)</math> | ||
| + | |||
| + | <math>r_1+r_2+r_3=-3</math> | ||
| + | |||
| + | <math>r_1r_2+r_2r_3+r_1r_3=9</math> | ||
| + | |||
| + | <math>r_1r_2r_3=-3</math> | ||
| + | |||
| + | In <math>g(x)</math> | ||
| + | |||
| + | <math>r_1+r_2+r_3+r_4=-4</math> | ||
| + | |||
| + | <math>=>r_4=-1</math> | ||
| + | |||
| + | <math>r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=6p</math> | ||
| + | |||
| + | <math>=>r_1r_2+r_1r_3+r_2r_3+r_4(r_1+r_2+r_3)=6p</math> | ||
| + | |||
| + | <math>=>9+(-1)(-3)=6p</math> | ||
| + | |||
| + | <math>=>p=2</math> | ||
| + | |||
| + | <math>r_1r_2r_3+r_1r_3r_4+r_1r_2r_4+r_2r_3r_4=-4q</math> | ||
| + | |||
| + | <math>=>-3+r_4(r_1r_3+r_1r_2+r_2r_3)=-4q</math> | ||
| + | |||
| + | <math>=>-3+(-1)(9)=-4q</math> | ||
| + | |||
| + | <math>=>q=3</math> | ||
| + | |||
| + | <math>r_1r_2r_3r_4=r</math> | ||
| + | |||
| + | <math>=>(-3)(-1)=r</math> | ||
| + | |||
| + | <math>=>r=3</math> | ||
| + | |||
| + | so <math>(p+q)r=\fbox{15}</math> | ||
| + | |||
| + | By ~Ahmed_Ashhab | ||
| + | |||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | Notice that to obtain the <math>x^4</math> term one must multiply <math>x^4+4x^3+6px^2+4qx+r</math> by some linear function of the form <math>x-a</math>. Looking at the <math>x^3</math> term, it is clear that <math>a</math> must equal <math>1</math>. Therefore by multiplying <math>x^4+4x^3+6px^2+4qx+r</math> by <math>x+1</math>, the product will be <math>x^4+4x^3+12x^2+12x+3</math>. Therefore <math>p=2</math>, <math>q=3</math>, <math>r=3</math>. Thus <math>(2+3)3=\fbox{15}</math> | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 40p box|year=1965|num-b=19|after=20}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 08:40, 26 November 2022
Contents
Problem 19
If 
is exactly divisible by 
, the value of 
is:
Solution 1
Let 
and 
.
Let 3 roots of 
be 
and 
. As 
, 3 roots of 4 roots of 
will be same as roots of . Let the 4th root of be 
. By vieta's formula
In
In
so 
By ~Ahmed_Ashhab
Solution 2
Notice that to obtain the 
term one must multiply 
by some linear function of the form 
. Looking at the 
term, it is clear that 
must equal 
. Therefore by multiplying by 
, the product will be 
. Therefore 
, 
, 
. Thus 
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
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