Question

1.) if x + 1 is a factor of ax^3 + 2x^2 - x + 3a - 7, then find the value of a.

2.) if p(x) = x^3 - 4x^2 + x + 6 then show that p(3) = 0 & hence factorise p(x).​

Answer 1

Q1)

Answer

➡

☑ the given equation is

(ax^3 + 2x^2 - x + 3a - 7 ) having factor (x + 1 )

(x + 2 ) is the factor of the equation therefor x = (- 1)

put x = -1 in the equation

(ax^3 + 2x^2 - x + 3a - 7 )

⟹(a(-1)^3 + 2(-1)^2-(-1)+3(a)-7

⟹ -a + (-2) +1+ (3a) - 7

⟹ -a - 2 +1+3a - 7

⟹ -a - 8 + 3a

⟹2a - 8

⟹a = 4

Verification

put the value of a and x in a given equation ,and equate the equation with zero

⟹(ax^3 + 2x^2 - x + 3a - 7 ) =0

⟹((4)(-1)^3+2(-1)^2-(-1)+3(4)-7=0

⟹((-4)+(-2)+1+12-7=0

⟹(-6+13-7)=0

⟹(-13+13)=0

⟹0=0

Hence verified

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Q2)

Answer

☑

given polynomial is

p(x) = x^3 - 4x^2 + x + 6

However,

we have to prove p(3)=0

put x = 3

p(3)=(3)^3-4(3)^2+(3)+6

⟹27-36+9

⟹27-27

⟹0

p(3)=0

Hence proved

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