1
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Question

# If (x+3) and (x+13) are factors of ax2+bx+5, then the value of 10a–3b+1 is equal to

A
1
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B
8
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C
4
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D
-2
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Solution

## The correct option is A 1Let f(x)=ax2+bx+5 It is given that (x+3) and x+13 are factors of f(x). By factor theorem, f(−3)=0 and f(−13)=0 ∴f(–3)=a(–3)2+b(–3)+5=0 ⇒9a−3b+5=0.........(i) f(−13)=a(−13)2+b(−13)+5=0 ⇒a9−b3+5=0 ⇒a−3b+45=0...........(ii) Subtracting (ii) from (i), we get: 8a – 40 = 0 ⇒a=5 …..(iii) Substituting a = 5 in (i), we get: 45 – 3b + 5 = 0 ⇒3b=50 ⇒b=503.........(iv) Using (iii) and (iv), we get: 10a – 3b + 1 =10(5)−3(503)+1 = 50 -50 + 1 = 1 Hence, the correct answer is option (1).

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