Trinomials of the Form ax² ± bx ± c
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Q.
Factorise: 2x2−56x+112
Q.
Factorize:
x2+6√2x+10
Q.
Factorise:
x2+1235x+135
Q. Factorize
x³—2x²—x+2
x³—2x²—x+2
Q.
Factorize:
2x2+3√5x+5
Q.
Factorize:
x2−2√2x−30
Q.
Factorize:
x2+5√5x+30
Q.
the factors of x3−7x+6 are
x9x−6(x−1)
(x2−6)(x−1)
(x+1)(x+2)(x−3)
(x−1)(x+3)(x−2)
Q.
Factorize:
x2+2√3x−24
Q.
Factorise:- 1-x3
Q.
Factorise:
21x2−2x+121
Q.
If (x-1) is a factor of x4+x3−2x2+x+1. Find the other factor using the factor theorem.
x3+2x2+x+1
x3+x2+x
x3+2x2+1
x3+2x2+1
Q. Factorize: 3a2−1−2a
- (3a−1)(a+1)
- (a−1)(3a+1)
- (3a+1)(a+1)
- (a−1)(3a−1)
Q.
Find the factors of x2+22x+112 by splitting the middle term.
(x+16)(x+3)
(x+9)(x+8)
(x+11)(x+8)
(x+14)(x+8)
Q.
Factorize:
x2−√3x−6
Q.
Factorize:
5√5x2+20x+3√5
Q.
4yz(z2+6z−16)÷2y(z+8) gives
z-2
z(z-2)
2z(z-2)
None of these
Q. Factorize: (2x+3y)2−12x−18y−7
- (2x+3y+1)(2x+3y−7)
- (2x+3y+1)(2x+3y+7)
- (2x+3y−1)(2x+3y−7)
- (2x+3y−1)(2x+3y+7)
Q. If (x+a)(x+b)=x2+5x+6, what is the value of (a−b) if (a−b)>0?
- 1
Q. Factorize 4x4−x2−12x−36
- (2x+3)(x+2)(2x2+x+6)
- (2x−3)(x−2)(2x2+x+6)
- =(2x+3)(x−2)(2x2+x+6)
- (2x+3)(x−2)(2x2−x−6)
Q. Factorize: 25x2−90xy+81y2
- (9x+5y)2
- (9x−5y)2
- (5x−9y)2
- (5x−9y)2
Q. ax2+bx+c can be factorised using the method of 'splitting the middle term' if b2−4ac is a perfect square.
- False
- True
Q. Factorise:
(i) x3−2x2−x+2
(ii) x3−3x2−9x−5
(iii) x3+13x2+32x+20
(iv) 2y3+y2−2y−1
[ 5 marks ]
(i) x3−2x2−x+2
(ii) x3−3x2−9x−5
(iii) x3+13x2+32x+20
(iv) 2y3+y2−2y−1
[ 5 marks ]
Q. Factorise 2(ab+cd)−a2−b2+c2+d2
- (c+d−a+b)(c+d−a−b)
- (c+d+a+b)(c+d−a−b)
- (c+d+a−b)(c−d−a−b)
- (c−d+a+b)(c−d−a−b)
Q.
Which of the following are the factors of 12ab+9a+12+16b?
(3a−4)and (4b−3)
(3a−4)and (4b+3)
- (3a+4)and (4b+3)
- (3a+4)(4b−3)
Q. Factorize: 6a2−a−15
- (2a−3)(3a−5)
- (2a+3)(5a−3)
- (2a+3)(3a−5)
- (3a+2)(2a−5)
Q. If x−2 is a factor of x2+3ax−2a, then a=
Q. If x−3 is a factor of x2−ax−15, then a=
Q. Factorize: 4x2−16xy
- 4x(x+4y)
- 2x(4x−y)
- x(4x−6y)
- 4x(x−4y)
Q. Factorize:
x2−14x+45
x2−14x+45