**Proof of this factor theorem**

Let

*p*(*x*) be a polynomial of degree greater than or equal to one and*a**be*areal number such that*p*(*a*) = 0. Then, we have to show that (*x*–*a*) is a factor of*p*(*x*).Let

*q*(*x*) be the quotient when*p*(*x*) is divided by (*x – a*).By remainder theorem

Dividend = Divisor x Quotient + Remainder

*p*(

*x*) = (

*x – a*) x

*q*(

*x*) +

*p*(

*a*) [Remainder theorem]

⇒ p(x) = (x – a) x q(x) [p(a) = 0]

⇒ (x – a) is a factor of p(x)

Conversely, let (

*x – a*) be a factor of*p*(*x*). Then we have to prove that*p*(*a*) = 0Now, (

⇒ p(x), when divided by (x – a) gives remainder zero. But, by the remainder theorem, p(x) when divided by (x – a) gives the remainder equal to p(a). ∴ p(a) = 0

*x – a*) is a factor of*p*(*x*)⇒ p(x), when divided by (x – a) gives remainder zero. But, by the remainder theorem, p(x) when divided by (x – a) gives the remainder equal to p(a). ∴ p(a) = 0

**Proof of remainder theorem.**

Let

*q*(*x*) be the quotient and*r*(*x*) be the remainder obtained when the polynomial*p*(*x*) is divided by (*x*–*a*).Then,

*p*(*x*) = (*x*–*a*)*q*(*x*) +*r*(*x*), where*r*(*x*) = 0 or some constant.Let

*r*(*x*) =*c*, where*c*is some constant. Then*p*(

*x*) = (

*x*–

*a*)

*q*(

*x*) +

*c*

Putting

p(a) = (a–a) q(a) + c ⇒ p(a) = 0 x q(a) + c ⇒ p(a) = c

*x*=*a*in*p*(*x*) = (*x*–*a*)*q*(*x*) +*c*, we getp(a) = (a–a) q(a) + c ⇒ p(a) = 0 x q(a) + c ⇒ p(a) = c

This shows that the remainder is

*p*(*a*) when*p*(*x*) is divided by (*x*–*a*).**Check Your understanding**

1. Factories

(i) a2-b2-4ac+4c2 (ii) 7x2 + 2 √14x + 2

(iii) 4a

^{2}-4b^{2}+4a+1 (iv) x^{4}+y^{4}-x^{2}y^{2}(v) x

^{6 }- x^{3 }(vi) x^{3}-5x^{2}-x+5(vii) x

^{2}+3√3x +6 (viii) a^{3}(b-c)^{3}+b^{3}(c-a)^{3}+c^{3}(a-b)^{3}(ix) .8a

^{3}-b^{3}-12a^{2}b +6ab^{2 }(x) b.4x^{2}+9y^{2}+ 25z^{2}-12xy - 30yz +20xz(xi)

*x*^{3}+ 4*x*^{2}+*x*– 6 (xii) 4x^{4}+ 7x^{2}– 2(xii) x

^{ 2 }- 2√3x – 45 (xiii) 3 - 12(a - b)2Q. Find degree of 5x

^{3 -}6x^{3}y+10y^{2}+11 [4]Q. find the value of k if(x-2) is a factor of p(x)=k x

^{2}^{-- }√2x +1 [ (2√2- 1)/4 ]Q. Find the remainder when x

^{3}+3x^{2}+3x+1 when divided by 3x+1.Q. if x-1 and x-3 are the factors of p(x) x (raise to the power 3)-a x (raise to the power 2)-13x-b then find the value of a and b

Q. If(X2-1) is a factor of ax

^{4}+bx^{3}+cx^{2}+dx+e,show that a + c + e = b + d =0Q. prove that (x+y)

^{3}-(x-y)^{3}-6y(x^{2}-y^{2})=8y^{3}Ans: x

^{3 }+ 3x^{2}y + 3xy^{2 }+ y^{3}- (x^{3 }- 3x^{2}y + 3xy^{2 }- y^{3}) - 6yx^{2 }+ 6y^{3}[*(a + b)*^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}*(a - b)*= x

^{3}= a^{3}- 3a^{2}b + 3 ab^{2}- b^{3}]^{3 }+ 3x

^{2}y + 3xy

^{2 }+ y

^{3}- x

^{3 }+ 3x

^{2}y - 3xy

^{2 }+ y

^{3 }- 6yx

^{2 }+ 6y

^{3 }= 2y

^{3 }+ 6 y

^{3 }= 8y

^{3}

Q. The polynomials {ax

^{3}-3x^{2}+4} and {3x^{2}-5x+a} when divided by {x-2} leave remainder "p" and "q" respectively. if p-2q+=a find the values of "a". [a is –8.]Q. If (

*x*− 4) is a factor of the polynomial 2*x*^{2}+*Ax*+ 12 and (*x*− 5) is a factor of the polynomial*x*^{3}− 7*x*^{2}+ 11*x*+*B*, then what is the value of (*A*− 2*B*)?Q. f x -1/x =3; then find the value of x

^{3}-1/x^{3 }[36]Q. if a+b+c=7 and ab+bc+ca=20 find the value of (a+b+c)

^{2}Q. The polynomial f(x)=x

^{4}-2x^{3}+3x^{2}-ax+b when divided by (x-1) and (x+1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x-2)Q. check: 2x +1 is a factor of p(x)=4x

^{3}+ 4x^{2}- x -1
x^4+1/x^4-3

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